CALCULUS 


F.  M.  AND  C. 


AN  ELEMENTARY  TREATISE 

ON 

CALCULUS 


A  TEXT  BOOK  FOR 
COLLEGES  AND  TECHNICAL  SCHOOLS 


WILLIAM  S.  FRANKLIN,  BARRY  MacNUTT 
AND  ROLLIN  L.  CHARLES 


OT   LEHIGH   XJNIVBHSITT 


SOUTH  BETHLEHEM,  PA. 
PUBLISHED  BY  THE  AUTHORS 

1913 

All  rights  reserved 


o(\ 


o 


03 


^7 


Copyright,  1913 
By  Franklin,  MacNutt  and  Charles 


Set  up  and  electrotyped.     Published  April,  1913 


PRESS  or 

THE  NEW  ERA  PRINTING  COMPANY 
LANCAbTER.  PA. 


PREFACE. 


1.  We  believe  that  the  most  important  thing  in  the  teaching  of 
calculus  is  to  lead  the  student  to  a  clear  understanding  of  principles. 
Therefore  our  chief  endeavor  has  been  to  develop  the  subject  as 
simply  and  as  directly  as  possible. 

2.  We  fully  appreciate  the  importance  of  extensive  practice  in 
the  handling  of  algebraic  transformations.  Therefore  we  have 
given  an  adequate  collection  of  formal  problems  in  differentiation 
and  integration. 

3.  We  are  convinced,  however,  that  it  is  a  pedagogical  mistake, 
in  a  text  book  on  calculus  for  young  men,  to  break  the  thread  of 
the  textual  discussion  by  unnecessary  algebraic  developments 
and  by  large  and  frequent  groups  of  purely  formal  problems. 
We  believe  indeed  that  the  proverbial  unintelligibility  of  calculus 
is  to  a  very  great  extent  a  psychological  consequence  of  this  almost 
universal  and  really  hideous  feature.  Therefore  we  have  arranged 
the  greater  portion  of  our  formal  problems  in  an  appendix. 

4.  Nearly  every  elementary  science  text  in  existence  carries  a 
false  suggestion  of  completeness  and  finality,  and  there  are  two 
things  which  young  men  should  understand  in  connection  with 
their  study  of  the  mathematical  sciences.  The  first  is  that  such 
study  is  exacting  beyond  all  compromise,  involving  as  it  does  a 
degree  of  constraint  which  it  is  beyond  the  power  of  any  teacher 
greatly  to  mitigate.  And  the  second  is  that  the  completest  science 
stands  abashed  before  the  infinitely  complicated  and  fluid  array 
of  phenomena  of  the  material  world,  except  only  in  the  assurance 
which  method  gives.  We  hope  that  this  elementary  treatise  on 
calculus  may  prove  to  be  sufficiently  definite  and  exacting  to  be 
useful;  next  to  this  there  is  nothing  we  coul3Vish  to  have  more  in 
evidence  from  beginning  to  end  than  its  incompleteness. 

273389 


VI  PREFACE. 

In  order  to  emphasize  the  incomplete  character  of  this  text  we 
have  introduced  references  to  more  complete  treatises  throughout 
the  text  and  we  have  given  in  Appendix  C  a  carefully  selected  list  of 
treatises  on  mathematics  and  on  mathematical  physics.  Teachers 
who  use  this  text  should,  we  believe,  direct  the  students*  attention 
to  this  appendix. 

In  our  brief  reference  to  the  infinitesimal  method  in  article  20 
we  do  not  wish  to  be  thought  of  as  taking  sides  in  the  old  contro- 
versy as  between  the  "  method  of  limits "  and  the  "  method  of 
infinitesimals"  in  calculus.  Article  20  is  unquestionably  falla- 
cious as  it  stands  ;  and  the  same  may  be  said  of  the  discussion  of 
divergence  and  curl  in  articles  126  and  129.  Indeed  articles  20, 
126  and  129  may  be  characterized  as  mere  plausibilities,  and  the 
harder  one  tries  to  understand  them  the  more  vague  and  unintel- 
ligible they  become.  The  fact  remains,  however,  that  the  infini- 
tesimal method  contributes  very  greatly  to  directness  and  sim- 
plicity of  speech  in  the  discussion  of  physical  problems,  and  the 
idea  of  infinitesimals  is  therefore  used  throughout  this  text.  Any 
one  who  is  disturbed  by  the  element  of  easy  plausibility  that  is 
involved  in  the  s  raight-forward  use  of  the  infinitesimal  method 
in  the  discussion  of  physical  problems  should  heed  the  advice 
given  by  D'Alembert  to  a  young  student,  "Go  ahead,  young  man, 
go  ahead!    Conviction  will  come  to  you  later." 

"The  absolute  requisites  for  the  study  of  this  work  are:  a 
knowledge  of  elementary  algebra  to  the  binomial  theorem  (accord- 
ing to  the  usual  arrangement),  plane  and  solid  geometry,  trigono- 
metry, and  the  most  simple  parts  of  the  usual  apphcations  of 
algebra  to  geometry." 

This  was  said  by  De  Morgan  in  the  preface  to  his  great  treatise 
on  Differential  and  Integral  Calculus  (London,  1842),  in  com- 
parison with  which  this  book  is  a  primer. 

Franklin,  MacNutt  and  Charles. 
South  Bethlehem,  Pa.,  March  22,  1913. 


CALCULUS. 


In  the  study  of  phenomena  which  depend  upon  conditions 
which  vary  in  time,  that  is,  upon  conditions  which  vary  from 
instant  to  instant,  it  is  necessary  to  direct  the  attention  to  what 
is  taking  place  at  an  instant;  or,  in  other  words,  to  direct  the 
attention  to  what  takes  place  during  a  very  short  interval  of 
time;  or,  borrowing  a  phrase  from  the  photographer,  to  make 
a  snap-shot,  as  it  were,  of  the  varying  conditions.  In  the  study 
of  phenomena  which  depend  upon  conditions  which  vary  from 
point  to  point  in  space,  the  attention  must  be  directed  to  what 
takes  place  in  a  very  small  region. 

This  paying  attention  to  what  takes  place  during  a  very  short 
interval  of  time  or  in  a  very  small  region  of  space  does  not  refer 
to  observation  but  to  thinking,  it  is  a  mathematical  method  and  it 
is  called  calculus.  Thus  the  density  of  a  body  is  its  mass  divided 
by  its  volume.  This  definition,  and  indeed  the  idea  of  density, 
appHes  only  to  a  homogeneous  substance.  To  apply  the  idea 
of  density  to  a  non-homogeneous  substance  one  must  think  of  a 
very  small  portion  of  the  substance.  The  same  thing  is  true 
of  every  measurable  property  of  a  substance.  Thus  the  idea  of 
elastic  strain  as  a  measurable  effect  is  easily  established  when 
every  part  of  a  body  is  similarly  strained  as  in  the  case  of  a 
stretched  rod,  but  to  apply  this  precise  idea  of  strain  to  a  bent 
beam  or  to  a  twisted  rod  one  must  think  of  a  small  portion  of 
the  beam  or  rod. 

Two  distinct  methods  are  involved  in  the  directing  of  the 
attention  to  what  takes  place  during  very  short  intervals  of  time 
or  in  very  small  regions  of  space,  as  follows: 

(a)  The  methgdj:iljii^[srmtial  cakiihisv  -  A  phenomenon  may  be 
prescribed  as  a  pure  assumption,  and  the  successive  instantaneous 


viii  CALCULUS. 

aspects  may  then  be  derived  from  this  prescription.  Thus  we 
may  prescribe  uniform  motion  of  a  particle  in  a  circular  path, 
and  then  determine  the  acceleration  of  the  particle  at  each  instant. 

Or,  a  condition  in  space  may  be  prescribed  as  a  pure  assumption, 
and  the  minute  aspects  may  then  be  derived  from  this  prescrip- 
tion. Thus  we  may  prescribe  the  distribution  of  temperature 
along  a  rod,  and  then  determine  the  temperature  gradient  at 
each  point  from  this  prescription. 

(6)  The  method  of  integral  calculus.  It  frequently  happens  that 
we  know  and  can  easily  formulate  the  action  which  takes  place 
at  a  given  instant  or  in  a  small  region  of  space.  The  problem 
then  is  to  build  up  an  idea  of  the  result  of  this  action  throughout 
a  finite  interval  of  time  or  throughout  a  finite  region  of  space. 
Thus  the  acceleration  of  a  body  may  be  known  at  each  instant, 
and  from  this  knowledge  we  may  find  the  velocity  gained  and 
the  distance  traveled  in  a  finite  interval  of  time.  Or,  consider  a 
disk  rotating  at  a  given  speed.  It  is  easy  to  establish  a  formula 
for  the  energy  of  a  very  small  particle  of  this  rotating  disk,  and 
it  is  then  possible  to  derive  an  expression  for  the  energy  of  the 
entire  disk. 


**% 


TABLE  OF  CONTENTS. 

PAGE 

Preface  and  Introduction v-x 

Chapter         I.     A  General  Survey  of  Differential  and 

Integral  Calculus 1-37 

Chapter       II.     Formulas  for  Differentiation  and  Inte- 
gration      38r  73 

Chapter  III.     Integration 74-  86 

Chapter  IV.  Partial  Differentiation  and  Integration    87-108 

Chapter  V.  Miscellaneous  Applications  of  Calculus  109-152 

Chapter  VI.     Expansions  in  Series 153-167 

Chapter  VII.  Some  Ordinary  Differential  Equations  168-189 

Chapter  VIII.     The  Partial  Differential  Equation  of 

Wave  Motion 190-209 

Chapter     IX.     Vector  Analysis 210-253 

Appendix  A.    Problems 1-20 

Appendix  B.    Table  of  Integrals 21-  28 

Appendix  C.    Some  Important  Books  on  Mathematical 

Theory 29-  38 


ix 


"  In  view  of  the  acknowledged  difficulty  of  calculus 

\the  student  must  be  willing  to  stop  in  his  course 

luntil  he  can  form  exact  notions  and  acquire  precise 

ndeas.^' 

Augustus  De  Morgan. 


CHAPTER  I. 

GENERAL  SURVEY  OF  DIFFERENTIAL  AND  INTEGRAL 
CALCULUS. 

1.  Constant  quantities  and  variable   quantities.— Elementary 

algebra  deals  only  wil^h  quantities  which  do  not  change  in  value 
(constant  quantities).  For  example  let  x  be  the  quantity  which 
subtracted  from  12  leaves  a  remainder  equal  to  Jx.  That  is, 
12  —  a;  =  ^x,  whence  x  =  S.  The  quantity  x  has  the  same 
value  throughout  this  discussion  and  it  is  therefore  called  a 
constant  quantity.  Calculus  is  a  branch  of  algebra  and  it  deals 
not  only  with  constant  quantities  but  also  and  especially  with 
quantities  which  change  in  value  {variable  quantities).  Thus  one 
of  the  simplest  problems  of  calculus  is  to  consider  how  rapidly  x^ 
grows  in  value  when  x  is  imagined  to  grow  at  a  definite  rate. 

Variable  quantities  fall  intQjthree^ fajxl^jdistinct  .classES,.  naiaely : 
(a)  Quantities  which  vary  in  time,  (h)  quantities  which  vary  in 
space,  and  (c)  quantities  which  are  arbitrarily  assumed  to  vary. 
An  example  of  the  first  class  is  the  varying  temperature  of  the  air 
at  a  given  place  as  the  seasons  come  and  go.  An  example  of  the 
second  class  is  the  varying  temperature  of  the  air  from  place  to 
place  at  a  given  time. 

Variables  of  the  first  and  second  classes  are  sometimes  called 
natural  variables  to  distinguish  them  from  variables  of  the  third 
class,  examples  of  which  are  considered  in  Arts.  10  and  11. 

2.  Value  at  a  given  instant.    Value  at  a  given  point. — Let    y 

represent  the  amount  of  water  in  a  pail  into  which  a  small  stream 
of  water  is  flowing.  Evidently  y  is  sl  changing  quantity.  If  the 
flow  of  water  were  to  be  shut  off  at  any  given  instant  there  would 
be  a  definite  quantity  of  water  left  in  the  pail,  and,  therefore,  there 
is  a  definite  amount  of  water  in  the  pail  at  each  instant  even  while 
2  1 


2  /'V  r  -'  :.•:».•    ^  .  ■  CALCULUS. 

the  inflow  continues.     Any  changing  quantity    y    has  a  definite 
value  at  each  instant* 

A  quantity  which  varies  in  space  has  a  definite  value  at  each 
point.*  Thus  everyone  understands  that  there  is  a  definite  tem- 
perature at  Philadelphia  although  it  may  be  different  from  the 
temperature  at  New  York  or  Washington.  An  iron  rod  with  one 
end  in  the  fire  has  a  definite  temperature  at  each  point. 

3.  Increments  and  decrements. — It  is  frequently  desirable  to 
consider  an  increase  (or  a  decrease)  in  the  value  of  a  variable 
quantity.  Such  an  increase  is  called  an  increment,  A  decrease  is 
called  a  decrement.  For  example,  let  y  he  &  variable  quantity. 
An  increment  of  y  is  usually  represented  by  the  symbol  Aj/. 
When  Ay  is  negative  it  is  a  decrement.  This  symbol  Ay  does 
NO  T  mean  A  multiplied  by  y,  it  is  a  single  symbol,  and  it  may 
be  read  increment-of-y  or  delta-y.  The  latter  is  preferable  because 
of  its  brevity.  The  Greek  letter  A  used  as  a  prefix  always 
means  increment  of. 

4.  Rate  of  change  of  a  quantity  which  varies  in  time. — Consider 
a  pail  into  which  water  is  flowing  in  a  small  stream.  Let  y  be 
the  amount  of  water  in  the  pail.  Then  y  is  evidently  a  changing 
quantity.  Consider  the  amount  of  water  that  flows  into  the  pail 
during  a  short  interval  of  time  At;  this  amount  of  water  is  evi- 
dently the  increment  of  y   during  the  short  intervaj.-of  timCj  and 

Ay 
it  is  to  be  represented  by  the  symbol    Ay.     The  quotient     -j 

At 

is  called  the  average  rate  of  change  of   y    duHng_ihe -interval — ^."f^ 

If  the  inflowing  stream  of  water  is  constant,  say,  always  2  cubic 

inches  of  water  per  second,  then  if  At  is  chosen  smaller  and  smaller, 

the  value  of   Ay   will  be  proportionately  smaller  and  smaller,  and 

Ay 
the  value  of  the  quotient    -^     will   be  exactly  2  cubic  inches 

per  second  whatever  the  duration  or  length  of  the  time-interval  A^. 

*  This  may  not  be  true  of  a  discontinuous  variable. 

t  Thus  if  y  is  the  amount  of  money  a  man  has  saved,  then,  if  he  saves 
$20  more  during  30  days,  we  have  Ay  =  $20  and  At  =  30  days,  and  the  aver- 
age rate  of  saving  during  the  30  days  will  be  66 1  cents  per  day. 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  3 

If  the  inflowing  stream  is  variable,  then  the  value  of  the  quotient 

~  will  not  be  the  same  for  different  values  of  Ai,  but  the  amount 

of  water  which  flows  into  the  pail  during  a  very  short  interval  of  time 
will  be  very  nearly  proportional  to  the  interval.  For  example, 
a  certain  amount  of  water  W  flows  into  the  pail  during  a  particular 
one-thousandth  of  a  second.  Imagine  this  particular  one-thou- 
sandth of  a  second  to  be  divided  in  halves;  then  only  a  very  Httle 
more  than  ^W  will  flow  into  the  pail  during  one  of  the  half- 
thousandths  of  a  second  and  a  very  httle  less  than  |TF  will  flow 
into  the  pail  during  the  other  half-thousandth  of  a  second.  // 
a  shorter  and  shorter  interval  of  time  he  taken  the  amount  of  inflow 
of  water  will  be  more  and  more  nearly  in  EXACT  proportion 
to  the  duration  of  the  interval.    This  means  that  the  quotient 

-Tj-  approaches  a  definite  limiting  value  as  At  and  Ay  both  ap- 

All 
proach  zero;  and  this  limiting  value  of   -rf    is  called  the  rate  of 

change  of  y  at  a  certain  instant  or  the  instantaneous  rate  of  change 

of  y. 

All  dii 

The  limiting  value  of  -^    is  always  represented  by    -^    which 

means  the  rate  of  change  of  y  at  a  given  instant. 

Nearly  everyone  falls  into  the  idea  that  such  an  expression  as 
10  feet  per  second  means  10  feet  of  actual  movement  in  an  actual 
second  of  time,  but  a  body  moving  at  a  velocity  of  10  feet  per 
second  might  not  continue  to  move  for  a  whole  second  or  its 
velocity  might  change  before  a  whole  second  has  clasped.  Three 
cubic  inches  per  second  is  the  same  rate  of  inflow  of  water  into  a 
pail  as  2,070  cubic  yards  per  year,  but  to  specify  the  rate  of  inflow 
in  cubic  yards  per  year  does  not  mean  that  a  whole  cubic  yard  of 
water  flows  into  the  pail  nor  does  it  mean  that  the  inflow  continues 
for  a  year.  A  man  does  not  need  to  work  for  a  whole  month  to 
earn  money  at  the  rate  of  $60.00  per  month,  nor  for  a  whole  day 
to  earn  money  at  the  rate  of  $2.00  per  day.     A  falling  body  has 


4  CALCULUS. 

a  velocity  of  19,130,000  miles  per  century  after  it  has  been  falling 
for  one  second,  but  to  specify  its  velocity  in  miles  per  century 
does  not  mean  that  it  falls  as  far  as  a  mile  nor  that  it  continues  to 
fall  for  a  century !  The  units  of  length  and  time  which  appear  in  the 
specification  of  a  velocity  are  completely  swallowed  up,  as  it  were, 
in  the  idea  of  velocity;  and  the  same  thing  is  true  of  the  specifica- 
tion of  any  rate. 

5.  Continuous  variables  and  discontinuous  variables. — A  quan- 
tity which  changes  by  sudden  jumps  is  called  a  discontinuous 
variable.  For  example  the  amount  of  money  one  has  is  a  dis- 
continuous variable,  because  a  debt  is  made  on  the  instant  that 
one  decides  to  accept  a  purchase;  that  is  to  say,  money  is  spent  in 
lumps,  and  a  lump  of  money  is  spent  during  an  indefinitely  short 
time.  The  amount  of  money  spent  during  an  interval  of  time  does 
NOT  become  more  and  more  nearly  proportional  to  the  interval  as  the 
interval  grows  shorter  and  shorter,  and  consequently  it  is  meaningless 
to  speak  of  the  rate  of  spending  money  at  a  given  instant.  If  y  is 
a  discontinuous  variable  and  if  the  time-interval  A^  happens  to 
include  a  jump  in  the  value  of  y,   then  the  value  of  Ay  remains 

finite  as  At  approaches  zero  and  the  quotient  -r^  approaches  in- 
finity. The  rate  of  change  of  a  discontinuous  variable  at  a  given 
instant  is  unthinkable. 

The  amount  of  water  in  a  pail,  as  considered  in  Art.  4,  is  an 
example  of  what  is  called  a  continuous  variable.  If  y  is  a  con- 
tinuous variable  and  if  Ay  is  the  increment  of  y  during  the  time- 
interval  At,  then  Ay  becomes  more  and  more  nearly  proportional 

to    At    as    A^    approaches  zero;  that  is  to  say,  the  quotient    -J 

a,pproaches  a  definite  limiting  value  as  At  approaches  zero,  and  this 

limiting  value  of  -~    is  called  the  instantaneous  rate  of  change 
zir 

of  y. 

In  the  study  of  calculus  we  deal  almost  exclusively  with  con- 
tinuous variables. 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  5 

6.  Example  showing  determination  of  instantaneous  rate  of 
change. — Let  us  assume  that  the  amount  of  water  in  the  pail  in 
the  discussion  of  Art.  4  is  proportional  to  the  square  of  the  time 
t  which  has  elapsed  since  a  chosen  instant.    Then  we  may  write 

y  =  kt'  (1) 

where  /:  is  a  constant.  A  moment  later  t  has  increased  and  of 
course  y  has  increased  also.  Let  us  represent  the  new  value  of  t 
by  t  +  At  and  the  new  value  of  y  by  y  -\-  Ay.  Then,  since 
equation  (1)  is  assumed  to  be  true  for  all  values  of  t  and  y,  we 

have 

y  +  Ay  =  k{t  +  Aty 
or 

y  +  Ay  =  kt^  +  2kt'At  +  h{AtY  (2) 

Subtracting  equation  (1)  from  equation  (2)  member  by  member, 

we  have: 

Ay  =  2kt'At  -f-  k{AtY  (3) 

Whence,  dividing  both  members  by  Aty  we  have: 

^  ^2kt  +  k'At  (4) 

Now  it  is  evident  that   k-At   approaches  zero  as   At   approaches 

Av 
zero,  and  therefore  the  limiting  value  of  -ry    (as   At   approaches 

dv  Av 

zero)  is    2kt.    Hence,  writing    -~    for  tY.e  limiting  value  of    -r^, 

we  have : 

|  =  2«  (5) 

That  is,  the  rate  of  change  of  y  is  at  each  instant  equal  to   2kt. 

Observation  or  thinking;  which? — The  discussion,  in  Art.  4, 
of  the  rate  of  increase  of  the  amount  of  water  in  a  pail  presents  a 
serious  difl&culty.  How  is  one  to  know  the  increment  of  water 
which  occurs  during  an  indefinitely  short  interval  of  time?    One 


6  CALCULUS. 

certainly  cannot  measure  it.  Observation  and  measurement  are 
entirely  usele!<s  in  the  consideration  of  an  indefinitely  small  change 
which  takes  place  during  an  indefinitely  short  interval  of  time. 
One  cannot  observe  such  things,  one  can  only  think  about  them. 
Indeed  the  whole  matter  at  present  under  discussion  is  a  matter 
of  mathematical  reasoning. 

7.  Gradient  of  a  quantity  which  varies  in  space. — Consider  an 
iron  rod  which  stands  with  one  end  in  a  fire.  Then  the  tempera- 
ture of  the  rod  varies  from  point  to  point  along  the  rod.  Consider 
two  points  very  near  together,  let  AT  be  their  difference  in 
temperature  and  let    Ax    be  their  distance  apart.     The  quotient 

AT 

— -   approaches  a  definite  limiting  value  when  Ax  is  chosen  smaller 

i\X 

AT 
and  smaller,  and  this  limiting  value  of  —  is  called  the  temperature 

gradient  at  the  point  (degrees  per  inch).  In  this  example  T 
varies  continuously  along  the  rod,  it  does  not  vary  by  jumps,  other- 
wise the  gradient  of   T  at  a  point  would  be  unthinkable. 

AT 
The  limiting  value  of  the  quotient  —  is  always  represented  by 

m^X 

dT 

j-j  and  it  means  the  gradient  of   T  at  a  given  point. 

Another  quantity  which  varies  from  point  to  point  in  space  is 
the  pressure  of  the  atmosphere;  the  higher  one  goes  the  less  the 
pressure.  Consider  two  points  A  and  B,  one  of  which  is  at  a 
distance  Ax  above  the  other,  and  let  Ap  be  the  amount  by  which 

the  pressure  at    A    exceeds  the  pressure  at    B.    Then  -—^   the 

Av 
limiting  vahie  of  the  quotient   -^,    is  called  the  pressure  gradient  at 

A.*  The  pressure  gradient  of  the  atmosphere  at  sea  level  is 
usually  about  2.3  pounds-per-square-inch  per  mile. 

The  word  gradient  as  used  in  the  above  discussion  comes  from 
the  word  grade  meaning  the  steepness  of  a  slope.     Thus  if  the 

*  Or  at  B)  the  two  points  A  and  B  approach  coincidence  as  Ax  ap- 
proaches zero. 


iJ&M^^ 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  7 

slope  of  a  hill  at  a  given  place  rises  A^  feet  in  a  horizontal  distance 

of  Aa;  feet,  then  the  limiting  value  of  —  when  Ax  is  made  smaller 

and  smaller  is  called  the  grade  of  the  hill  at  the  given  place. 
The  grade  of  a  hill  at  a  point  might  be  one  foot  of  rise  per  ten 
feet  of  horizontal  distance,  or  ten  feet  of  rise  per  hundred  feet  of 
horizontal  distance,  or  10  per  cent,  as  it  is  usually  expressed. 
It  is  evident  that  10  feet  of  rise  per  100  feet  of  horizontal  distance 
does  not  mean  necessarily  10  feet  of  actual  rise  in  100  feet  of  hori- 
zontal distance,  because  the  slope  may  not  be  100  feet  long  and 
the  grade  may  vary  from  point  to  point. 

Similarly  a  pressure  gradient  of  2.3  pounds-per-square-inch  per 
mile,  which  is  the  upward  gradient  of  atmospheric  pressure  at  sea 
level,  does  not  mean  that  the  atmospheric  pressure  at  a  point  one 
mile  above  sea  level  is  2.3  pounds  per  square  inch  less  than  at  sea 
level,  because  the  pressure  gradient  becomes  less  and  less  at  points 
higher  and  higher  above  the  sea.  The  units  which  appear  in  the 
specification  of  a  grade  or  a  gradient  are  completely  swallowed  up, 
as  it  were,  in  the  idea  of  grade  or  gradient  just  as  the  units  which 
appear  in  the  specification  of  a  rate  are  swallowed  up  in  the  idea  of 
rate. 

8.  Example  showing  the  determination  of  a  gradient  at  a  point. 

— Consider  a  metal  bar    AB,    Fig.  1.    Suppose  the  temperature 

e6td T h&i 

k X  — -^ 

Fig.  1. 

of  the  bar  to  be  zero  at  the  end  A,  and  let  us  assume  the  tempera- 
ture  T  at  the  point  p  to  be: 

T  =  hx"  (1) 

where  x  is  the  distance  from  A  to  p,  and  A;  is  a  constant. 
Under  these  conditions  let  it  be  required  to  find  the  temperature 
gradient  at  the  point  p.  Now  equation  (1)  is  true  for  every  value 
of    X    and    T.     Therefore  writing    x  -\-  Ax    for    x    and  writing 


8  CALCULUS. 

T  +  AT  for   T  we  have 

T  +  AT  =  ik(x  +  Ax)*  (2) 

or 

T  +  AT  =  A:^*  +  2fcc-Aa;  +  A:(Ax)*  (3) 

Subtracting  equation  (1)  from  equation  (3)  member  by  member, 

we  have: 

AT  =  2ib:-A:t  +  A:(Ax)*  (4) 

whence 

^  =  2A:x  +  k-^x  (5) 

Now    A; -Ax    approaches  zero  as    Ax    approaches  zero,  and  it  is 

AT 
therefore  evident  that  the  Umiting  value  of  —  is  2kx,    Therefore, 

/iX 

writing  -r-   for  the  limiting  value  of  -— ,  we  have: 

g  =  2fcx  (6) 

It  is  evident  from  this  equation  that  the  temperature  gradient  is 

zero  at  the  end  A  of  the  bar  where  x  =  0;  and  it  is  evident  that 

the  temperature  gradient  grows  greater  and  greater  (steeper  and 

steeper)  at  points  farther  and  farther  from   A   where  x  is  larger 

.  and  larger. 
J  PROBLEMS. 

1.  The  amount  of  water  in  a  pail  is  assumed  to  be  proportional 
to  the  square  of  the  time  which  has  clasped  since  a  chosen  instant, 
according  to  equation  (1)  of  Art.  6.  It  is  evident  from  this  equa- 
tion that  the  amount  of  water  in  the  pail  is  zero  when  i  =  0. 
After  the  lapse  of  20  seconds  there  is  600  cubic  inches  of  water  in 
the  pail.  Find  the  value  of  k  and  state  the  unit  in  terms  of  which 
k  is  expressed.     Ans.  1.5  cubic  inches  per  second  per  second. 

2.  Using  the  value  of  k  from  problem  1,  find  the  rate  at  which 
water  flows  into  the  pail,  (a)  when  t  =  5  seconds,  and  (6)  when 
t  =  20  seconds.  Ans.  (a)  15  cubic  inches  per  second,  (6)  60 
cubic  inches  per  second. 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  9 

3.  Find  the  average  rate  of  inflow  of  water  into  the  pail  during 
the  time  from  t  =^  0  to  t  =  20  seconds.  Ans.  30  cubic  inches  per 
second. 

4.  The  temperature  of  an  iron  rod  increases  along  the  rod  in 
accordance  with  equation  (1)  of  Art.  8.     At  a  point  20  inches  from 
the  end  of  the  rod  (x  =  20  inches),  the  value  of    jT    is  240°  C.   ; 
Find  the  value  of    k    and  state  the  unit  in  terms  of  which   it    is 
expressed.     Ans.  0.6  degree  per  inch  per  inch. 

5.  Using  the  value  of  k  from  problem  4,  find  the  temperature 
gradient  (a)  at  the  point  where  x  =  5  inches  and  (b)  at  the  point 
where  x  =  20  inches.  Ans.  (a)  ^  degrees  per  inch,  (6)  24  d^rees 
p>er  inch. 

6.  (o)  Find  the  average  temperature  gradient  along  the  iron 
rod  of  problem  4  between  x  =  0  and  x  =  20  inches.  (6)  Find 
the  average  temperature  gradient  between  x  =  10  inches  and 
X  =  11  inches.  Ans.  (a)  12  d^;rees  per  inch,  (6)  12.6  degrees  per 
inch. 

9.  Arbitrary  variations. — One  of  the  most  important  things  in 
calculus  is  to  consider  how  fast  such  an  expression  as  x*  or  x*  or 
sin  X  changes  when  x  is  arbitrarily  assumed  to  grow  steadily  in 
value.  Such  -variations  may. be  called  arbitrary  variations.  The 
ver>'  great  importance  of  arbitrary-  variations  will  be  understood 
when  the  student  reaches  Arts.  22  and  23. 

10.  A  purely  algebraic  example  of  an  arbitrary  variation. — 
Having  given  the  equation 

y  =  aa^  (1) 

let  it  be  required  to  find  the  limiting  value  of  the  quotient    — 

when  the  arbitrary  increment  of  x,  namely  Ax,  approaches  zero. 
Writing  y  -{-  Ay  for  y  and  writing  x  +  Ax  for  x  in  equation 
(1)  we  have 

y  +  Ay  =  a(x  +  Ax)*  (2) 

or 

y  +  Ay  =  ax»  4-  2axAx  +  a(Ax)*  (3) 

whence,  subtracting  equation  (1)  from  equation  (3)  member  by 


10  CALCULUS. 

memoer,  we  have 

Ai/ =  2aa:-Aa;  +  a(Ax)«  (4) 

and  dividing  both  members  by  Ax  we  have 

^  =  2ax  +  a-Ax  (6) 

Ax 

But  a 'Ax  approaches  zero  when  Ax  approaches  zero,  and  there- 
fore it  is  evident  that  the  limiting  value  of    -r^    is    2ax.    The 

Ax 

.  Ay  dv 

limitmg  value  of  -r^   is  always  represented  by  ~.    Therefore  we 
i^c  dx 

have: 

g  =  ^  (6) 

11.  Two  more  purely  algebraic  examples  of  arbitrary  variations. 

— (a)  Having  given  the  equation 

y  =  as?  (1) 

let  it  be  required  to  find  the  limiting  value  of  -^  when  the  arbi- 
trary increment  of  x  approaches  zero.  Writing  y  -\-  Ay  for  y 
and  writing  x  +  Ax  for  a;  in  equation  (1)  we  have 

y-^Ay  =  a{x  +  AxY  (2) 

or 

y-\-  Ay  ^  a3?  -\-  Sax^-Ax  +  3ax(Ax)2  +  a{^xY  (3) 

whence,  subtracting  equation  (1)  from  equation  (3)  member  by 
member,  we  have 

Ay  =  3ax2.Ax  +  3ax(Ax)2  +  a(Ax)«  (4) 

or 

^  =  3ax2  +  3ax-Ax  +  a{Axy  (5) 

Now  3ax-Ax  and  a{Axy  both  approach  zero  when  Ax  approaches 

Ay 
eero,  and  therefore  it  is  evident  that  the  limiting  value  of   ~    is 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  11 

Sax^;  that  is, 


#  _ 


dx 
(6)  Having  the  equation 


3ax2  (6) 


y  =  l  (7) 

let  it  be  required  to  find  the  limiting  value  of  ~  when  the  arbi- 
trary increment  of  x  approaches  zero.  Writing  y  -\-  Ay  for  y 
and  writing  x  +  Ax  for  x  in  equation  (7)  we  have 

y  +  ^y-^^  («) 

Subtracting  equation  (7)  from  equation  (8)  member  by  member, 
we  have 

A2/  =  -^  -  ?  (9) 

^       x  +  Ax       X 

Reducing  the  two  fractions  to  a  common  denominator  we  have: 


whence 

Ay 


(11) 


Ax            x^  -\-  x-Ax 
but  x^  -\-  X'Ax  approaches  x'^   as  its  limit  when  Ax  approaches 
zero,  and  therefore  the  limiting  value  of   ■—  is ^  ;  that  is : 

dx  X'  ^^^^ 

dy 
The  derivative  —   is  sometimes  called  the  rate  of  change  of  y 

with  respect  to  x. 


12  CALCULUS. 

PROBLEMS. 

The  following  problems  can  be  solved  by  the  methods  employed 
in  Arts.  10  and  11.  In  each  case  £uid  the  rate  of  change  of  y 
with  respect  to  x. 

l.y^ax,  g  =  a. 

2.  2/  =  ax»,  ^  =  5ax^, 

3.  y  =  ms?  +  w,  -p  =  27nx. 

4.  2/  =  oar^  +  6a;  +  c,  ^  =  2ax  +  6. 

5.  ^  =  ax^  +  6x2,  ^  =  3ac'  +  2&a;. 
-            a                         di/  2a 

_       g  dy  _  —  a(2hx  +  1) 

**^~6a;'»  +  x'  5i 

n     10.2/  =  =^,  * 

V  '^      6  —  x'  da; 

11.     y    ^    2X^-5  ^ 


12.  1/ 


X  dy  ^ 

(x-l)2'  d!x  {x-iy 


(6X2  +  3.)2 

9 

(x  +  sr 

ab 

(6  -  x)2- 

9 

(x  +  2)2- 

x  +  1 

12.  Functions. — When  a  spring  is  subjected  to  a  stretching  force 
Ff  a  certain  elongation    e    is  produced  as  shown  in  Fig.  2.     For 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS. 


13 


every  value  of  e  there  is  a  definite  corresponding  value  of  F.    When 
two  quantities  are  associated  in  this  way  either  one  is  said  to  be  a 


un  stretched  spring 


Fig.  2. 
The  elongation  of  the  spring  is  a  function  of  the  stretching  force. 

function  of  the  other.  Thus  the  elongation  e  produced  by  a 
given  force  F  in  Fig.  2  is  a  function  of  F;  or  the  force  F  re- 
quired to  produce  a  given  elongation   e   is  a  function  of  e.    The 


1 

1 

K -^  i j| 

w 

1 
1 
1 
1 

area^lw 

Fig.  3. 

The  area  of  a  square  is 
function  of  length  of  side. 


Fig.  4. 

The  area  of  a  rectangle  is  a 
function  of  length  and  width. 


meaning  of  the  word  function  is  further  illustrated  by  Figs.  3, 4, 5, 
6,  7  and  8. 

13.  Dependent  and  independent  variables. — If  I  in  Fig.  3  is 
assumed  to  be  arbitrarily  variable  it  is  called  an  independent 
variable,  and  the  area  of  the  square  is  called  a  dependent  variable 
because  its  value  is  determined  or  fixed  when  the  value  of  I  is 
given.  The  quantity  of  air  pumped  into  the  cylinder  in  Fig.  7 
may  be  as  much  as  one  pleases,  the  temperature  of  the  cylinder 


14 


.    CALCULUS. 


and  air  may  be  increased  or  decreased  at  will,  and  the  volume  of 
the  cylinder  may  be  changed  by  moving  the  piston.    Therefore 

tr- thermometer 


^ 


"l-cloaed  • 
'. :  vessel  •'•' 
containing 
':  air .  •  - 


Fig.  5. 

The  pressure  of  the  air  is  a  function  of 
the  amount  of  air  pumped  into  the  tire. 


Fig.  6. 

rhe  pressure  of  the  air  in  the 
vessel  is  a  function  of  the  amount 
of  au-  pumped  into  the  vessel  and  of 
the  temperature. 


the  three  quantities,  (a)  quantity  of  air,  (6)  temperature  of  the  air, 
and  (c)  volume  of  space  occupied  by  the  air  are  called  independent 
variables,  and  the  pressure  of  the  air  is  called  a  dependent  variable, 

thermometer 

pressure 
gauge 


piston 


cylinder 
containing  \  air 


SiLnp 


Fig.  7. 

The  pressure  of  the  air  in  the  cylinder  is  a  function  of  (a)  the  amount  of 
air  pumped  into  the  cylinder,  (6)  the  temperature,  and  (c)  the  volume  of 
the  space  occupied  by  the  air. 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS. 


15 


given  curve 


because  its  value  is  fixed  or  determined  when  the  quantities  (a), 
(6)  and  (c)  are  given. 

From  this  discussion  it  is  evident  that  a  dependent  variable 
may  be  a  function  of  one,  two,  three  or  more  independent  variables. 

14.  Natural  functions  and  artificial  functions. — We  have  hereto- 
fore taken  the  point  of  view  that  two  quantities  must  be  connected 
physically  if  either  is  to  be  a  function  of  the  other.  Such  a  function 
may  be  called  a  natural  func- 
tion. It  is,  however,  of  very 
great  importance  to  consider  func- 
tional relations  which  grow  out 
of  Or  are  expressed  by  algebraic 
equations.  Such  functions  may 
be  called  artificial  functions.  Some 
idea  of  the  importance  of  artificial 
functions  may  be  obtained  by 
looking  back  at  Articles  4  and  6, 
and  7  and  8.  Natural  functions 
can  be  in  many  cases  expressed 
algebraically^;  indeed  no  function  can  be  handled  in  calculus  unless 
the  function  is  expressed  algebraically. 

15.  Tabulation  of  a  function. — ^When  one  quantity  is  a  function 
of  another  it  is  often  convenient  to  express  the  functional  relation 
by  means  of  a  table  giving  pairs  of  corresponding  values  of  the 
two  quantities.  An  ordinary  table  of  logarithms  is  such  a  table, 
a  table  of  sines  or  cosines  is  such  a  table,  the  mortality  table* 
used  in  life  insurance  is  such  a  table. 

When  a  functional  relation  is  determined  by  experiment  the 

*  If  one  were  to  consider  all  men  in  the  United  States  who  are  now  forty 
years  of  age  and  keep  a  future  record  of  their  deaths  one  would  find  that  their 
average  age  at  death  would  be,  say,  sixty-five  years.  ^This  is  expressed  by 
saying  that  the  expectancy  of  fife  at  forty  years  is  twenty-five  years,  because 
on  the  average  a  man  forty  years  old  may  expect  to  live  twenty-five  years. 
The  expectancy  of  life  is,  of  course,  less  and  less  with  increasing  age,  and  there 
is  a  definite  expectancy  of  life  corresponding  to  each  age.  I  That  is,  expectancy 
of  life  is  a  function  of  age.  7 


Fig.  8. 

The  ordinate  of  a  point  on  a  curve  is 
a  function  of  the  abscissa  of  the  point. 


16 


CALCULUS. 


Fig.  9. 


observed  quantities  are  always 
arranged  in  tabular  form.  For 
example  Fig.  9  represents  an  am- 
meter arranged  to  measure  the 
electric  current  flowing  through  a 
coil  of  wire,  and  a  spring  scale  ar- 
ranged to  measure  the  force  with 
which  an  iron  plunger  is  pulled  into 
the  coil;  and  the  accompanying 
table  shows  a  series  of  observed 
values  of  current  in  amperes  and  the 
corresponding  pulls  in  pounds. 


OBSERVED  RELATION  BETWEEN  CURRENT  AND  PULL  IN 

FIG.  9. 


Current  in 

Pull  on  Plunger 

Amperes. 

in  Pounds. 

1.0 

3.9 

2.0 

23.6 

S.0 

40.0 

10 

48.1 

a. 


16.  Graphical  representation  of  a  fimction. — The  relation  be- 
tween the  pull  of  the  spring 
and  the  reading  of  the  am- 
meter in  Fig.  9,  as  shown  in 
the  table  of  Art.  15,  may  be 
represented  graphically  by 
a  curve  of  which  the  ab- 
scissas represent  ammeter 
readings  and  of  which  the 
ordinates  represent  corre- 
sponding readings  of  the 
spring  scale.  Such  a  curve 
is  shown  in  Fig.  10.  In  the 
same  way  an  algebraic  func- 
tion may  be  represented  graphically, 
represents  the  function  y  =  x^. 


50 
40 
30 

20 

10 

^ 

^ 

/^ 

/ 

/ 

/ 

/ 

^ 

[/ 

' 

345 

amperes 


Fig.  10. 
Thus  the  curve  in  Fig.  11 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS. 


17 


17.  Derivative  of  a  function. 

let  A?/  be  the  increment  of 
y  due  to  an  arbitrary  incre- 
ment of  X.  Then  the  lim- 
iting value  of  the  quotient 

-^  as    Ax  approaches  zero 

is  called  the  derivative  of  y 
mth  respect  to  x  or  the  rate 
of  change  of  y  with  respect 
to   X* 

When  one  wishes  to  speak 
in  general  terms  of  any  func- 
tion of  X  and  its  derivative, 
the  function  may  be  repre- 
sented by  f{x)  or  by  <^(x) 
and  the  derivative  of  the 


-Let   2/   be  a  function  of   x   and 


I  X'OxiiB 


Fig.  11. 


function  with  respect  to  x  may  be  represented  hy  f'{x)  or  by  <l>'{x). 

Meaning  of  a  derivative. — Let  i/  be  a  function  of  x  and  let   x 

dy 
be  assumed  to  grow  steadily  at  a  definite  rate,  then   y    grows  -r- 

times  as  fast  as  x  at  each  instant.     For  example,  let  y  =  x^;  then 

dif 

~  =  2x,    according  to  Art.  10,  and  the  following  relations  exist: 


y    increases  2x{=  20)    times  as  fast  as    x 

value  X  —  10, 
y   increases   2a; (=  22)    times  as  fast  as   x 

value  X  =  11, 
y  increases   2a:  (=  24)    times  as  fast  as  x 

value  X  —  12, 
^  etc.,  etc., 


while  ; 
while  ; 
while   ; 

As  another  example,  let  y  =  -;  then  -^ 

X  dx 


is  passing  through  the 
is  passing  through  the 
is  passing  through  the 


etc. 

1_ 

x^' 


according  to 


Art.  11,  and  the  following  relations  exist: 

*  Sometimes  also  called  the  differential  coefficient  of    y    with  respect  to    x. 


18 


CALCULUS. 


y  decreases  ^  (  ~  4  )    as  fast  as  x  increases  when  x  is  passing  through  the 


value  2, 

y  decreases    -j  (   =  qJ  as  fast  as  x  increases  when  x  is  passing  through  the 

value  3, 

etc.,  etc.,  etc.  --> 

A  negative  value  of  ^  shows  that  y  decreases  as  x  increases. 
ax 

The  derivative  of  a  function  of    x    is  itself  a  function  of    x. 

— If  y  =  dx^.  then  ^  =  2ax,  according  to  Art.  10.     In  this  case 
ax 

dti 
it  is  evident  that  -p  is  a  function  of  x  because  it  is  equal  to  2ax 

and  it  has  a  definite  value  for  every  value  of  x.  If  /(x)  represents 
any  function  of  x,  then  its  derivative  }'{x)  is  in  general  a  function 
of  X  also. 

18.  Graphical  representation  of  derivative.    Slope  of  a  curve 
at  a  point. — Consider  the  curve  CCy  Fig.  12.    This  curve  represents 


Fig.  12. 

a  definite  function,  that  is,  the  ordinate  ?/  is  a  definite  function  of 
the  abscissa  x,  the  point  p  being  anywhere  on  the  curve.  If  x 
is  increased  by  the  amount  Ax,  then   y  will  be  increased  by  the 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  19 

amount  A^  as  shown,  and  the  ratio  -f-  is  equal  to  the  tangent  of 

the  angle   a. 

Draw  the  hne  tt  touching  the  curve  at  p  as  shown.  As  Aa; 
approaches  zero  the  angle  a  becomes  more  and  more  nearly 
equal  to  the  angle    6,   in  fact   6  is  the  limiting  value  of  a,  and, 

of  course,  the  limiting  value  of    -r^     is      ~.    Therefore,  since 

tan  a  =  — ^  ,  we  must  have  tan  ^  =  -r^-    That  is,  the  function   y 

dti 
being  represented  by  the  ordinates  of  a  curve,  the  derivative    -r-    is 

represented  by  the  steepness  or  grade  of  the  curve  at  each  point. 

19.  Derivative  notation  and  differential  notation. — Consider  the 
function 

y  =  ax^  (1) 

of  which  the  derivative  is: 

I  =  ^  (2) 

dv 
Heretofore  we  have  looked  upon  ^-  as  a  single  symbol  the  mean- 
ing of  which  is  explained  in  Art.  17.    Convenience  of  notation 
sometimes  makes  it  desirable  to  write  equation  (2)  thus: 

dy  =  2ax'dx  (3) 

This  equation  expresses  the  limiting  relation  between  the  incre- 
ments A^  and  Ax,  that  is  to  say,  the  relation  which  is  approached 
as  Ay  and  Ax  both  approach  zero;  dy  is  called  the  differential  of 
y  (that  is  to  say,  the  differential  of  ax^,  because  y  here  stands  for 
ax^),  and  dx  is  called  the  differential  of  x.  These  two  differentials 
may  be  thought  of  as  indefinitely  small  increments  of  y  and  x 
respectively.* 

*  When  y  =  ax^  we  have 

Ay  =  2ax  •  Ax  +  ^(Ax)^ 

according  to  Art.  10.  That  is  to  say,  the  increment  of  y  is  not  equal  to 
2ax  times  the  increment  of  x  unless  both  increments  are  indefinitely  smaU. 


20 


CALCULUS. 


Or,  if  one  wishes  to  think  of  dy   and   dx   as  having  a  physical 


dt 


meaning  they  may  be  thought  of  as  abbreviated  expressions  for 

dx 
and  -rr  respectively,  that  is,  dy  may  be  thought  of  as  the  rate  of 

change  of  y  and  dx  may  be  thought  of  as  the  rate  of  change  of  x, 

PROBLEMS. 

1.  A  pail  is  10  inches  in  diameter,  so  that  the  volume  of  water  in 

the  pail  is  y  =  —^  Xj  where  x  is  the  depth  of  the  water  in  the 

pail.  Find  how  fast  water  must  be  poured  into  the  pail  to  cause 
the  water  level  to  rise  at  a  velocity  of  4  inches  per  second.  Ans. 
IOOtt  cubic  inches  per  second. 

*/t2.  Water  flows  at  a  constant  rate  of  30  cubic  inches  per  second 
into  a  metal  cone  of  which  the  dimensions  are  as  shown  in  Fig.  p2. 
Find  the  velocity  at  which  the  water  level  rises 
(a)  when  x  =  b  inches,  and  (6)  when  a;  =  15 
inches.  Ans.  (a)  2.715  inches  per  second,  (6) 
0.302  inch  per  second.  €m  t'c--       ^ 

3.  The  sides  of  a  square  are  growing  at  the 
rate  of  5  inches  per  second.  Find  the  rate  of 
growth  of  the  area  of  ^  the  square  (a)  when 
the  sides  of  the  square  are  10  inches  and  (6) 
when  the  sides  of  the  square  are  20  inches. 
Ans.  (a)  100  square  inches  per  second,  (6)  200 
square  inches  per  second. 

4.  The  radius  of  a  circle  is  growing  at  the 
rate  of  5  inches  per  second.  Find  the  rate  of 
growth  of  the  area  of  the  circle  (a)  when  the 

radius  is  10  inches  and  (h)  when  the  radius  is  20  inches.  Ans.  (a) 
lOOir  square  inches  per  second,  (h)  2007r  square  inches  per  second. 

5.  The  edges  of  a  cube  are  growing  at  the  rate  of  5  inches  per 
second.  Find  the  rate  of  growth  of  the  volume  of  the  cube  when 
the  edges  are  10  inches  long.     Ans.  1,500  cubic  inches  per  second. 


Fig.  p2. 


DIFFERENTIAL  i^fe  INTEGRAL  CALCULUS.  21 


m 

9  of  iR] 


-  6.  Air  is  blown  into  a  soapi«3ble  at  the  rate  of  10  cubic  inches 
per  second.  Find  the  rate  of  Slrease  of  the  radius  of  the  bubble 
when  the  radius  has  the  values  (^  1  inch  and  (6)  4  inches.  Ans. 
(a)  0.795  inch  per  second,  (6)  0.050  inch  per  second. 

-7.  A  man  6  feet  tall  walks  at  a  speed  of  4  miles  per  hour  under 
a  lamp  which  is  10  feetfrgmthe  ground.     Find  how  fast  ^€-tip  • 
ef-  the  man's  shadow  wa^eS-mien  the  horizontal  distance  from 
the  lamp  to  the  man  is  20  feet.     Ans.  6  miles  per  hour. 

8.  The  two  sides  of  a  right  angled  triangle  are  40  inches  and  x^ 
and  the  hypothenuse  is  y.  The  side  z  is  growing  at  the  rate  of 
5  inches  per  second.  How  fast  is  y  growing  when  x  =  30  inches? 
Ans.  3  inches  per  second. 

9.  The  two  sides  of  a  right  angled  triangle  are  x  and  y,  and 
the  hypothenuse  is  50  inches.  The  side  x  is  growing  at  the  rate 
of  4  inches  per  second.  How  fast  is  y  growing  (a)  when  x  =  30 
inches,  and  (6)  when  x  =  40  inches.  Ans.  (a)  3  inches  per 
second.     (6)  5.33  inches  per  second. 

10.  The  observed  temperatures  of  a  vessel  of  cooling  water  after 
1  minute,  after  2  minutes,  and  so  forth  are  as  follows: 

Elapsed  Time  in  Minutes,  ObserTed  Temperaturea, 

t  T. 

0  92.0 

1  85.3 

2  79.5 

3  74.5 
5  67.0 
7  60.5 

10  53.5 

15  45.0 

20  39.5 

Plot  these  values  of  t  and  T,  using  the  best  grade  of  squared 
paper,  draw  a  smooth  curve  through  the  plotted  points,  draw 
tangents  to  the  curve  at  the  points  corresponding  to  (a)  T  =  80°,, 
(h)  T  =  65°,  and  (c)  T  =  50°,  and  determine  the  rate  of  decrease 
of  the  temperature  at  each  point  by  measuring  the  intercepts; 
of  these  tangents  on  the  axes  of  reference. 


22  CALCULUS. 

11.  Find  for  what  values  of  x  the  function  ?/  =  Gx^  —  12x  is 
an  increasing  function  and  for  what  values  of  x  it  is  a  decreasing 
function.  Ans.  2/  is  an  increasing  function  when  x  is  greater 
than  1,  and  a  decreasing  function  when  x  is  less  than  1. 

12.  Plot  the  values  of  x  and  y^  where  y  =  x^,  using  the  best 
grade  of  squared  paper,  draw  a  smooth  curve  through  the  plotted 
points,  draw  tangents  to  the  curve  at  the  points  corresponding  to 
X  =  1,     x  =  2    and    x  =  3,     and  determine  the  corresponding 

values  of   -p   by  measuring  the  intercepts  of  the  tangents  on  the 

dii 
axes  of  reference.    Compare  the  values  of    -—    thus  found  with 

dxi 
the  values  as  calculated  from  the  formula  -—  =  Sx^. 

ox 

13.  Plot  the  values  of  x  and  y  where  y  =  logio  x.    Take  the 

values  of  logio  x  from  an  ordinary  table  of  logarithms  and  use  the 

following  values  of  x:  2,  3,  4,  5,  6,  7,  8,  9  and  10.     Draw  a  smooth 

curve  through  the  plotted  points,  draw  tangents  to  the  curve  at 

the  points  corresponding  to  x  =  4,  x  =  6  and  x  =  8,    and  deter- 

dv 
mine  the  corresponding  values  of  -^   by  measuring  the  intercepts 

of  these  tangents  on  the  axes  of  reference.     Compare  the  values  of 

dti 

-^     thus  found  with 

dx 

dy  ^  logioe  ^  0.4343  ^ 

dx  X  X     ' 

20.  Determination  of  limiting  relation  between  Ay  and  Ax  by 

Aiy 
consideration  of  infinitesimals. — The  value  of    -~    is  found  in 

Ax 

Arts.  10  and  11  as  the  sum  of  a  finite  qvxintity  and  a  quantity  which 

approaches  zero  as  Ax   approaches  zero.     Thus  when   y  =  as?  we 

found  that    -r^  =  Zax^  +  3ax-Ax  +  a(Ax)2,    in  which    Sax^    is  a 

ZaX 

finite  quantity  and     3ax-Ax  +  a(Ax)^     is  a  quantity  which  ap- 
proaches zero  as   Ax    approaches  zero.     In  such  a  case  it  is  very 


dti 

-^     thus  found  with  the  values  as  calculated  by  the  formula 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.      23 

A?/ 
easy  to  see  what  the  limiting  value  of  -~-   must  be. 

In  many  cases,  >oweve%  it  is  desirable  to  find  the  limiting  rela- 
tion between    ^y    and    Ax    when    Ax    approaches  zero  without 

deriving  an  expression  for  ~ .     For  example  let  y  =  a^.    Then 

from  Art.  11  we  have:  -  -   -        -^  ^ 

_/■'  ^ 
,  Ay  =  Sax^'Ax  +  Zax{Axy  +  a{Axy  (1) 

Both  members  of  this  equation  approach  zero  when  Ax  approaches 
zero,  and  it  would  therefore  seem  rather  difficult  to  determine 
the  hmiting  relation  between  Ay  and  Ax  (when  Ax  approaches 
zero)  from  this  equation  as  it  stands.  But  when  Ax  is  made  as 
small  as  you  please,  thejM|(Aa;)2  is  infinitely  smaller  than  Ax, 
and  {AxY  is  infi]||||fily  sn^^r  than  (Ax)^.  For  example  let  Ax 
be  a  millionth  of^Mpait^i^K  (Ax)^  is  a  million-millionth  of  a 
unit,  and  (Ax)^  is^3fcpioB;^Blion-millionth  of  aunit.  Therefore 
the  terms  3ax  •  (Ax)^and '  ^J\x)^  become  more  and  more  nearly 
negligible  in  comparison  with  Sax^  •  Ax  as  Ax  grows  smaller 
and  smaller  in  equation  (1),  The  limiting  relation  between  Ay 
and  Ax  may  be  found  by  writing  dx  and  dy  for  Ax  and  Ay 
to  indicate  that  we  have  proceeded  to  the  limit,  and  by  dropping 
every  term  which  contains  the  square  or  any  higher  power  of 
Ax    (or  the  square  or  any  higher  power  of   Ay).     This  gives 

dy  =  Sax^-c^x  (2) 

In  this  equation  dy  and  dx  are  as  small  as  you  please  and  they 
are  called  infinitesimals;  but  {dyY  and  {dxY  being  infinitely 
smaller  than  dy  and  dx  are  called  infinitesimals  of  the  second 
order  J  {dyY  and  {dxY  being  infinitely  smaller  than  {dyY  and 
{dxY  are  called  infinitesimals  of  the  third  order,  and  so  on.  In  any 
differential  expression  the  lowest  order  infinitesimals,  only,  are 
significant;  all  terms  containing  higher  order  infinitesimals  as  factors 
may  he  dropped. 


24 


CALCULUS. 


Example  1. — Let  it  be  required  to  differentiate  the  expression 

2/2  =  ax»  (3) 

that  is  to  say,  let  it  be  required  to  determine  the  relation  between 
Ay  and  Ax  when  Ax  is  as  small  as  you  please  Writing  {y  -{•  Ay) 
for  y  and  writing   (x  +  Ax)   for  x,   we  have 

y^-\-2yAy-i-  {Ay)^  =  ax^  +  Sax^-Ax  +  3ax(Ax)2  +  a(Ax)»    (4) 

whence,  subtracting  equation  (3)  from  equation  (4)  member  by 
member,  we  have 


2y'Ay  -\-  (Ay)^  =  3ax2-Ax  +  3ax(Ax)2  -f-  a{Ax)^ 


(5) 


from  which  the  desired  limiting  relation  is  found  by  dropping 
second  and  third  order  infinitesimals,  so  that  we  have: 


2y-dy  =  dax^-dx 


(6) 


The  value  of  y  from  equation  (3)  may  be  substituted  in  equation 
(6),  giving: 

2^[a^-dy  =  Zax^'dx  (7) 


which  can  be  simplified  by  cancellation,  giving: 
dy  =  j-^ax-dx 


(8) 


This  dropping  of  higher  order  infinitesimals  from  differential  expres- 
sions does  not  lead  to  merely 
approximate  results,  because  in 
every  case  it  is  the  limiting  rela- 
tion between  Ay  and  Ax  which 
is  to  be  determined,  that  is,  the 
relation  when  Ay  and  Ax  are  both 
as  small  as  you  please.  Equation 
(8),  for  example,  is  rigorously  true. 

Example  2.  Differential  of  the 
arc  of  a  parabola. — The  length  s  of 
the  heavy  portion  of  the  curve  cc, 


'X^' 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS. 


25) 


Fig.  13,  from  x  =  b  tox  =  x  is  a  function  of  x,  and  from  the  in- 
finitesimal triangle  whose  sides  are  dx,  dy  and  ds  we  have 


ds  =  ^l{dxy  +  {dyy 


or 


*=  >r+  ity-^ 


(9) 
(10) 


For  example  suppose  the  curve  cc,  Fig.  13,  to  be  a  parabola  whose 
equation  is: 

y  =  kx^  (11) 

then 


dy  _ 


dx 


=  2kx 


which,  substituted  in  equation  10,  gives: 


ds  =  ^ll-\-4:k^x^'dx 


(12) 


(13) 


21.  Functions  which  have  the  same  derivative. — The  curve   A, 
Fig.  14,  defines  i/'  as  a  function  of  a;,  and  the  curve  B  defines  y 


y-axiB 


Fig.  14. 

as  a  function  of  x.    The  steepness  or  grade  of  curve  A  is  every- 
where the  same  as  the  steepness  or  grade  of  curve  B,  that  is,  the 

derivative  -j-  is  equal  to  the  derivative  —  for  each  value  of  a;, 


26  CALCULUS, 

or  expressed  as  an  equation  we  have 

dx       dx 


(1) 


From  the  figure  it  is  evident  that  the  difference  1/  —  y  is 
everywhere  equal  to  the  constant  quantity  C.  Consequently 
when  equation  (1)  is  true  we  have 

y'  —  y  =  a.  constant  (2) 

That  is  to  say,  two  functions  whose  derivatives  are  equal  have  a 
constant  difference. 

Examples. — ^When  two  men  save  money  at  the  same  rate  there 
is  a  constant  difference  between  the  amounts  they  have  saved; 
or,  if  they  start  even,  their  savings  are  equal.  When  two  trains 
travel  at  the  same  speed  they  remain  at  a  constant  distance  apart. 
The  two  functions  ax^  and  ax^  +  h  have  the  same  derivative 
with  respect  to  x,   h  being  a  constant. 

22.  Example  showing  the  use  of  calculus.    Work  required  to 

stretch  a  spring. — Let  W  be  the  work  required  to  stretch  a  spring 

from  condition  A    to  condition  B,   Fig.  2.     It  is  evident  that   W 

is  a  function  of  e,  and  it  is  desired  to  find  an  algebraic  expression 

for  this  function;  that  is,  it  is  desired  to  find  an  equation  expressing 

W  in  terms  of  e.     To  do  this  the  first  step  is  to  find  an  expression  for 

dW 
the  derivative  -t~    by  considering  the  amount  of  work   ATF  that 

"^  "'-         /^ 

must  be  done  to  produce  a  slight  additional  elongation  Ae. 

The  force  F  in  Fig.  2  is,  according  to  Hooke's  law,  proportional 
to  e;  that  is, 

F  =  ke  (1) 

where  A;  is  a  constant. 

Imagine  the  force  F  to  be  increased  by  the  amount  AF  so 
that  the  elongation  would  be  increased  by  the  amount  Ae. 
Then  the  work  done  would  be  greater  than  F-Ae  and  less  than 
{F  -\-  AF)-Ae    because  the  average  value  of  the  force  which  acts 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  27 

on  the  spring  while  the  added  elongation  Ae   is  being  produced  is 
greater  than  F  and  less  than  F  +  AF.    Therefore 

AW  is  greater  than    F  •  Ae  (2) 

and 

AW  is  less  than   (F  -\- AF)  -  Ae  (3) 

whence,  dividing  by  Ae,  we  have : 

aw 

-^  is  greater  than  F  and  less  than   F  +  AF  (4) 

Ae 

AW 
Therefore  ——    approaches  F  as  its  limit  as  AF   (and  also  Ae) 

approaches  zero.     That  is, 

dW 


or,  using  the  value  of  F  from  equation  (1),  we  have 

f=^  ,,, 

Now  in  Art.  10  it  is  shown  that  ax^  is  a  function  whose  derivative 
is  2ax.    Therefore: 

ae2    is  a  function  whose  derivative  is    2ae  (7) 

Therefore,  substituting    k  =  2a    (or   a  =  \k)    in  (7)  we  have: 

\ke^  is  a  function  whose  derivative  is  ke  (8) 

The  work  W  is  also  a  function  whose  derivative  is  ke,  according 
to  equation  (6).     Therefore,  according  to  Art.  21,  we  must  have 

W  =  W  +  C  (9) 

where  C  is  a  constant.  To  determine  the  value  of  the  constant 
C  we  must  know  the  value  of  W  for  some  particular  value  of  e, 
and  of  course  we  know  that  TT  =  0  when  e  =  0,  that  is,  no  work 
is  required  when  the  spring  is  not  stretched  at  all.     Therefore, 


28 


CALCULUS. 


substituting  this  pair  of  values  of   W   and   e   in  equation  (9),  we 
have 

0  =  C  (10) 

Substituting  this  value  of    C  in  equation  (9)  we  have 

W  =  ike''  (11) 

which  is  the  desired  expression  for  the  work   W. 

23.  Another  example  showing  the  use  of  calculus.  The  area 
under  a  parabola. — The  curve  cc.  Fig.  15,  represents  a  parabola 
of  which  the  equation  is: 

y  =  px^  (1) 

and  it  is  required  to  find  an  expression  for  the  area  A. 
Imagine  the  value  of  x  in  Fig.  15  to  increase  by  the  amount  Ax^ 


X 

X  — 

Fig  15.  Fig.  16. 

then  the  increment  of  A  is  the  area  of  the  narrow  strip  in  Fig.  16. 
The  width  of  this  strip  is  Ax  and  its  height  is  y  (=  px^).    There- 


fore 
whence 


A  A  =  px^'Ax 
AA 


(2) 
(3) 


Now  the  width   Ax  in  Fig.  16  must  be  infinitely  small  in  order 
that  the  height  of  the  strip  may  be  considered  to  be  equal  to 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  29 

y  {=  px^) .    That  is,  we  have  already  proceeded  to  the  limit  in  writing 

down  equation  (2),  and  it  is  evident  that  equation  (3)  gives  the 

AA 
limiting  value  of  — .    Therefore  we  have 

In  Art.  11  it  was  shown  that: 

ac^   is  a  function  whose  derivative  is   Soi*  (5) 

Therefore  substituting  ^  for  a  we  have : 

Ipa^  is  a  function  whose  derivative  is  px'^  (6) 

The  area  A  is  also  a  function  whose  derivative  with  respect  to 
X  is  px^  according  to  equation  (4),  and  therefore,  according  to 
Art.  21,  we  must  have: 

A  =  ip:x^+C  (7) 

where  C  is  a  constant.  To  determine  the  value  of  the  constant 
C  we  must  know  the  value  of  A  for  some  particular  value  of  x. 
An  inspection  of  Fig.  15  shows  that  A  =  0  when  x  =  L  There- 
fore, substituting  this  pair  of  values  of  A  and  x  in  equation  (7) 
we  have: 

0  =  ipP  +  C  (8) 

so  that 

C  =  -  ipP  (9) 

Substituting  this  value  of    C  in  equation  (7)  we  have: 

A  =  ipa^  -  \pP  (10) 

which  is  the  desired  expression  for  the  area   A   in  Fig.  15. 

It  is  to  be  noted  that  a^  is  a  variable  in  the  above  discussion 
because  we  have  arbitrarily  made  it  vary  in  the  derivation  of 
equation  (4),  whereas  p  and   I  are  both  constant. 


30  CALCULUS. 

PROBLEMS. 

In  each  of  the  first  six  following  problems,  find  the  rate  of 
change  of  y  with  respect  to  x. 


1.  y^  =  3x, 

2.  y^  =  ax^  —  3x, 

3.  y^  =  2ax^  +  3x, 

4.  y^  =  a^  -  x\ 


dx      2  >x* 
d^  _  3(ax^  -  1) 
d^  ~X^ax^~^Zx ' 
dy  _      4:ax  +  3 
dx  ~i  (2aa;2  +  3x)« 


5    .2  =  1  ^=-,J. 

^*  ^       x'  <te  2a;f 

°'  ^       x'  dx  3x1* 

7.  The  volume  V  of  a  cone  of  which  the  half-angle  at  the  apex 
is  30°  is  a  function  of  the  altitude    x    of    the  cone.    Set  up  the 

derivative  -j-,     Ans.   -j-  =  \irx^. 

8.  Set  up  the  derivative  of  the  area  A  of  the  curved  surface  of 

dA 
the  cone  of  problem  7  with  respect  to  x.     Ans.    y-  =  ^irx. 

9.  The  volume  F  of  a  cone  of  which  the  half-angle  at  the  apex 
is  30°  is  a  function  of  the  slant  height  x  of  the  cone.     Set  up  the 

,    .     ,.       dV      .         dV       V3      2 
derivative  -,-.     Ans.   -5-  =  -^-ttx^. 
ox  ax         o 

10.  The  kinetic  energy    TT    of  a  disk  of  steel  rotating  at    n 

revolutions  per  second  is  a  function  of  the  radius   r    of  the  disk. 

dW 
Set  up  the  derivative    -7- ,    the  thickness  of  the  disk  being  10 
dr 

dW 
centimeters.    Ans.  -3—  =  787rnV 
dr 


■m^ 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  31 

Note. — The  density  of  steel  is  7.8  grams  per  cubic  centiraeter  and  the 
kinetic  energy  of  a  particle  in  ergs  is  \mv^  where  m  is  the  mass  of  the  particle 
in  grams  and  v  is  the  velocity  of  the  particle  in  centimeters  per  second. 

11.  The  shaded  area  in  Fig.  pll  is  a  function  of  the  angle  6. 
Find  the  derivative  of  the  area  with  respect  to  B,  the  angle  0  being 
expressed  in  radios  and  the  equation  of  the  curve  cc  being 
r  =  kB  -\-hf  where  k  and  h  are  constants. 

Ans.  ^  =  \{kB  +  by. 

12.  The  length  of  the  arc    s    in  Fig.  pl2  is  a  function  of  the 


Fig.  pll.  Fig.  pl2. 

angle  B.     Find  the  derivative  of  s  with  respect  to  B,  the  equation 

d  s  I 

of  the  curve  cc  being  as  given  in  problem  11.     Ans.  -^  =  V  fc^  -f  ^^. 

au 

24.  The  differential  equation.    Law  of  growth  of  a  function. — 

Let  y  be  an  undetermined  function  of  x  concerning  which  it  is 
known  that  y  increases  2ax  times  as  fast  as  x.  This  property 
of  the  unknown  function  may  be  expressed  by  the  equation: 

I  =  2«  « 

This  equation  expresses  the  law  of  growth  of  the  function  y,  the 
independent  variable  x  being  assumed  to  grow  steadily  in  value. 
Any  equation  which  expresses  the  law  of  growth  of  a  function  is 
called  a  differential  equation. 

Examples,  (a)  A  man  A  saves  money  at  a  constant  rate.* 
This  statement  is  a  verbal  expression  of  a  differential  equation. 

*  Money  saved  is  here  assumed  to  be  a  continuous  variable.    See  Art.  5. 


32  CALCULUS. 

To  reduce  the  statement  to  algebra,  let  y  be  the  amount  of  money 
saved  by  A,  then  the  above  statement  is  equivalent  to 

f  =  .  (2) 

where  A;  is  a  constant. 

(6)  One  man  A  saves  money  twice  as  fast  as  another  man  B, 
This  statement  also  is  a  verbal  expression  of  a  differential  equation. 
To  reduce  it  to  algebra  let  y  be  the  amount  of  money  saved  by  A 
and  let  x  be  the  amount  of  money  saved  by  B.  Then  while  B 
saves   Ax   dollars    A    must  save  twice  as  much  according  to  the 

above  statement.    Therefore  ^y  —  2Ax  or  -r^  =  2,  or 

i=2  (3) 

(c)  Equation  (6),  Art.  22,  is  a  differential  equation,  and  equation 
(4),  Art.  23,  is  a  differential  equation. 

25.  Differentiation  and  integration. — The  finding  of  the  deriva- 
tive of  a  function  as  exemplified  in  Arts.  10  and  11  is  called  dif- 
ferentiation. 

Let  y  be  an  unknown  function  of  x  which  is  known  to  in- 
crease 2ax  times  as  fast  as  x.  That  is  to  say,  it  is  known  concern- 
ing y  that 

1  =  20.  (1) 

Now  when  one  has  previously  found  that  ax^  is  a  function  of  x 
whose  derivative  is  eqvxil  to  2ax  one  is  able  to  recognize  that  the 
unknown  function  y  whose  law  of  growth  is  given  by  equation  (1) 
must  be  equal  to  ax^  -f  C,  where  C  is  a  constant,  as  explained  in 
Art.  21.  This  recognition  of  a  function  from  its  derivative  is 
called  integration. 
Symbol  of  integration. — Let  y  stand  for  ax^  +  C,  then 

dy 


dx 


2ax  (2a) 


.^vV 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  33 

that  is  to  say, 

the  derivative  of  (ax^  +  C)   is  equal  to  2ax  (26) 

Or,  using  the  differential  notation  which  is  explained  in  Art.  19, 
we  have 

dy  =  2aX'dx  (2c) 

that  is  to  say, 

the  differential  of  (ax^  +  C)   is  equal  to  2ax  •  dx         (2d) 

All  of  these  equations  (2a),  (26),  (2c)  and  {2d)  express  the  same 
thing  or  the  same  relation,  and  this  relation  may  be  expressed  in 
an  inverse  manner  by  saying: 

(ax^  +  C)  is  the  integral  of  2aX'dx  (3a) 

or,  in  symboUc  notation: 

ax'^-  C  =  J2ax'dx  (36) 

The  symbol  J  means  integral  of.  The  undetermined  constant 
C  is  called  the  constant  of  integration. 

A  differential  equation  does  not  completely  determine  the  func- 
tion whose  law  of  growth  it  expresses. — This  is  evident  when  we 
consider  that  the  function  y  -{-  C  has  the  same  derivative  (the 
same  law  of  growth)  whatever  the  value  of  the  constant  C  may  be. 

Thus  the  amount  of  a  man's  savings  cannot  be  calculated  when 
his  rate  of  saving  alone  is  known;  one  must  know  also  how  much 
money  the  man  had  on  a  given  date.  Thus  a  man  who  had  $1,000 
on  January  1,  1912,  and  who  saves  $10  per  month  would  have 
1,000  +  10m  dollars  at  any  time  m  months  after  January  1. 
The  amount  of  a  man's  savings  is  completely  determined  as  a 
function  of  elapsed  time  by  knowing  (a)  his  rate  of  saving  and 
(6)  how  much  he  had  on  a  given  date. 

In  general  a  function  y  is  completely  determined*  by  knowing 

*  This  statement  will  have  to  be  modified  when  we  come  to  consider  dif- 
ferential equations  of  higher  orders,  and  when  we  come  to  consider  partial 
differential  equations. 
4 


34  CALCULUS. 

(o)  its  law  of  growth  and  (b)  its  value  corresponding  to  any  given 
value  of  the  independent  variable  x  or  <  as  the  case  may  be. 
This  matter  is  fully  illustrated  in  Arts.  22  and  23. 

26.  Indefinite  integrals  and  definite  integrals. — Let    y    be  a 

dv 
function  of  x  concerning  which  nothing  is  known  except  that  -^ 

is  equal,  say,  to  2ax;   or,  using  differential  notation,  we  know  that 

dy  =  2ax'dx  (1) 

Then  according  to  Arts.  10  and  21  we  know  that  y  must  be  given 
by  the  equation: 

2/  =  0x2  +  C  (2) 

where  C  is  a  constant  which  may  have  any  value  whatever.  This 
equation  (2)  expresses  what  is  called  the  indefinite  integral  of  equa- 
tion (1). 

Let  it  he  required  to  find  the  growth  of  y  while  x  grows  from  any 
given  value  I  to  any  other  given  value  L.  This  growth  of  y  is 
called  a  definite  integral  of  equation  (1),  and  the  two  given  values 
of   X   are  called  the  boundaries  or  limits*  of  the  definite  integral. 

Now,  according  to  equation  (2),  we  have  y  =  aP  -]-  C  when 
X  =  I,  and  y  =  aL^  -{•  C  when  x  =  L;  and  the  difference 
between  these  two  values  of  y,  namely,  aU  —  al^,  is  the  desired 
growth  of  y  from  x  =  I  to  x  =  L.  That  is,  the  definite  integral 
of  equation  (1)  between  the  boundaries  or  Hmits  x  =  I  and  x  =  L 
is  equal  to  aL^  —  aP. 

Examples  of  definite  integrals. — Consider  the  area  A  under 
the  parabola  in  Fig.  15.     The  differential  of  this  area  is 

dA  =  px^'dx  (3) 

and  the  value  of  the  area  is 

A  =  ips^-  JpP  (4) 

*  The  word  limit  as  here  used  has  a  very  di£ferent  meaning  from  the  word 
limit  as  used  in  Arts.  4  and  7, 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  35 

as  explained  in  Art.  23.  Now  equation  (4)  is  the  definite  integral 
of  equation  (3)  from  x  =  I  to  x  =  x  (meaning  any  value  of  x 
whatever). 

Sjrmbolic  representation   of   a   definite   integral. — The  usual 
method  of  expressing  a  definite  integral  is 


=  j^'^px^'dx  (5) 


where  A  is  the  area  under  the  parabola  in  Fig.  15,  the  differential 
of  A  being  given  by  equation  (3)  above. 

Similarly,  the  work  W  required  to  stretch  the  spring  in  Fig.  2 
is  found  as  a  definite  integral  in  Art.  22,  and  to  express  W  sym- 
bolically as  a  definite  integral  we  write 


W^f' 


ke-de  j[6) 


The  value  of  this  definite  integral  is   Jfce^  as  explained  in  Art.  22. 

27.  A  definite  integral  interpreted  as  the  limit  of  a  sum. — Imagine  the 
area  A  of  Fig.  15  to  be  divided  into  narrow  strips  as  shown  in  Fig.  17,  the 
width  of  successive  strips  being  represented  by  successive  increments  Ax 
of  X,  starting  at  x  =  I  and  extending  to  any  given  value  of  x.  This  given 
value  of  X  is  represented  by  L  in  Fig.  17,  because  it  is  now  desirable  to 
think  of  X  as  having  various  intermediate  values  (between  x  =  I  and 
x  =  L). 

The  width  of  any  strip  is  Ax  and  its  altitude  is  px^,  where  x  is  the 
distance  of  the  particular  strip  from  the  y-axis.  Therefore  the  area  of  the 
strip  is  px^  •  Ax  and  the  sum  of  the  areas  of  all  the  strips  is  more  and  more 
nearly  equal  to  the  area  A  of  Fig.  15  as  the  number  of  the  strips  is  made  greater 
and  greater. 

The  area  A  is  equal  to    J^px^  •  dx  as  explained  in  Art.  26,  and  the  sum 

L 

of  all  the  strips  in  Fig.  17  may  be  represented  symbolically  as   ^px^  •  Ax. 
When  the  number  of  strips  is  made  greater  and  greater  (successive  increments 

L  PL 

Ax  made  smaller  and  smaller)  then    "^px^  •  Ax   approaches     f ,  px^  •  dx   as 

I  * 

.    .         ^  .     C^ 

a  limit.    That  is,  the  limit  of  ^px^  'Ax  is  J.  px^  •  dx. 


36 


CALCULUS. 


28.  Plan  of  this  treatise. — The  application  of  calculus  involves 

three  important  things.    Thus  in  Art.  22  the  following  three  things 

dW 
are  involved:  (a)  The  finding  of  the  derivative  -t-  by  making  use 

of  fundamental  principles  in  physics,  (6)  the  becoming  familiar 
with  the  derivative  of  ax^  in  Art.  10,  and  (c)  the  substitution  of 
Jfc  for  a  and  e  for  z  so  as  to  get  a  known  function  whose  deriva- 
tive is  ke. 

The  application  of  calculus 

in  engineering  involves  these 
three  steps  in  nearly  every 
case,  and  the  following  devel- 
opment of  calculus  consists  of 
three  parts  (more  or  less  inter- 
mingled) as  follows: 

(o)  The  setting  up  of  de- 
rivative expressions  (differen- 
tial equations)  with  the  help 
of  fundamental  principles  in 
physics. 

(b)  The  study  of  derivatives 
of  algebraic  functions;  and 
(c)  Practice  in  transforming  derivative  expressions  to  standard 
forms  for  purposes  of  recognition. 

The  most  important  branches  of  calculus  which  do  not  fall  into 
this  outline  are  the  discussion  of  maximum  and  minimum  values  of 
functions  and  the  expansion  of  functions  in  series 

A  table  of  functions  Q,nd  their  derivatives*  is  given  in  Appendix  B. 
The  student  should  get  into  the  habit  of  referring  to  this  table, 

PROBLEMS. 

1.  A  force  of  100  pounds  produces  4  inches  of  elongation  of  a 
spring.  Find  the  value  of  k  in  equation  (1)  of  Article  22,  and 
express  the  unit  in  terms  of  which  k  is  found.  Ans.  25  pounds 
per  inch. 

*  Or  of  differential  expressions  and  their  integrals. 


DIFFERENTIAL  AND  INTEGRAL  CALCULUS.  37 

2.  Using  the  spring  referred  to  in  problem  1,  find  the  work 
required  to  produce  (a)  an  elongation  of  2  inches,  and  (6)  an  elonga- 
tion of  4  inches,    Ans.  (a)  50  pound-inches,  (6)  200  pound-inches. 

3.  Using  the  spring  referred  to  in  problem  1,  find  how  much  work 
is  required  to  increase  the  elongation  from  3  inches  to  4  inches. 
Ans.  87.5  pound-inches. 

4.  Consider  the  parabola  y  =  2x^.  Find  the  area  under  this 
curve  between  ordinates  at  a;  =  2   and  x  =  10.    Ans.  66^J. 

5.  Find  an  expression  for  the  area  under  the  curve  y  =  qx 
between  an  ordinate  at  x  =  a  and  the  ordinate  corresponding  to 
any  abscissa  x.    Ans.  A  =  ^qx^  —  fga^. 


CHAPTER  II. 
FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION. 

29.  Differentiation  by  development  and  differentiation  by  rule. 
— Throughout  the  previous  chapter  the  derivative  of  a  function 
(as  in  Arts.  10  and  11)  or  the  differential  of  a  function  (as  in  Art. 

20)  is  determined  by  developing  an  expression  for   — ^  or  for  Ay, 

Ax 

and  considering  the  limiting  form  of  this  expression  as  Ay  and 
Ax  approach  zero.  This  is  the  fundamental  method  of  differentia- 
tion. It  is  possible,  however,  by  means  of  this  fundamental  method 
to  establish  a  few  rules  which  enable  one  to  write  down  the  derivative 
or  differential  of  almost  any  algebraic  function.  It  is  the  object  of 
this  chapter  to  establish  these  rules  and  give  examples  of  their  use. 
Differentiation  may  be  defined  in  general  as  the  finding  of  the 
rate  of  change  of  a  function.  Thus  problem  10  on  page  21  illus- 
trates what  may  be  called  graphical  differentiation.  And  to  de- 
termine a  rate  by  actually  measuring  the  change  which,  takes 
place  in  a  given  time  may  be  called  physical  differentiation.  Thus 
the  speed  of  a  runner  is  determined  by  measuring  the  distance 
traveled  by  the  runner  in  a  given  time,  or  rather  by  measuring 
the  time  required  for  the  runner  to  travel  a  given  distance. 
To  determine  the  rate  of  supply  of  water  by  a  small  spring  is 
to  "differentiate  the  spring"  and  the  simplest  method  is  to 
make  the  differentiation  by  means  of  a  dipper  and  a  dollar  watch. 

30.  Differentiation  of  ax". — Let 

y  =  ax"  (1) 

Then 

dy  =  nax"-i-dx  *  (2) 

for  any  value  of    n.    We  shall  here  prove  equation  (2)  when    n 
is  a  positive  integer.* 

*A  general  proof  is  given  in  Art.  39. 

38 


£J^^ 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  39 

Proof. — Writing  y  +  Ay  for  y  and  writing  x  +  Ax  for  x  in 
equation  (1)  we  have: 

y  +  Ay^a{x  +  Ax^  (3) 

When  n  is  a  positive  integer  {x  +  AxY  can  be  expanded  by  the 
binomial  theorem,*  and  we  have 

y  -\-  Ay  =  ax""  +  nax""-^  •  Ax  -\ — ^ ^^ — ^ — -  +  etc.    (4) 

Subtracting  equation  (1)  from  equation  (4),  member  by  member, 
we  have: 

Ay  =  nax^'-^'Ax  +  -^ ^ — - — -  +  etc.  (5) 

J.  *  z 

Whence,  writing  dy  for  Ay  and  dx  for  Ax,  and  dropping  in- 
finitesimals of  second  and  higher  orders,  we  have : 

dy  =  nax''~^'dx  (2) 

31.  Differentiation  of  the  sum  of  two  or  more  functions. — Let 

u  and  V  be  any  two  functions  of  x,   and  let 

y  =  u  +  V  (1) 

Then 

dy  =  du  +  dv  (2) 

or,  using  derivative  notation, 

dx      dx  ^  dx  ^^^ 

Proof. — It  is  evident  that  the  increment  of  y  is  equal  to  the 
sum  of  the  increments  of  u  and  v,  that  is.  Ay  =  Au  -\-  Av;  and 
this  is  true  however  small  the  increments  may  be.  Therefore 
when  the  increments  are  infinitesimal  we  have     dy  —  du  ■\-  dv. 

*  The  binomial  theorem  is: 

(a  +  6)«  =  a«  +  -J-  + — + —— +  etc. 

When  n  is  not  an  integer  this  becomes  an  infinite  series,  and  an  infinite  series 
cannot  be  used  in  a  mathematical  argument  unless  the  question  of  its  con- 
vergence is  carefully  considered.     See  Art.  35. 


40  CALCULUS. 

Also  it  is  evident  that 

Ay»_AuAv 
Ax~  Ax      Ax 

and  this  is  true  however  small  the  increments  may  be.    Therefore 
when  the  increments  are  infinitesimal  we  have 

d]i  _  du      dv 
dx      dx      dx 

Explanation. — The  fact  which  is  expressed  by  equation  (3)  is 
extremely  simple  and  it  is  self-evident  when  it  is  expressed  in 
familiar  terms  as  follows:  The  amount  of  money  earned  by  one 

man  is   u   and  his  rate  of  earning  is  2  dollars  per  day  (  =  75: )  J 

the  amount  of  money  earned  by  another  man  is   v   and  his  rate 

of  earning  is  3  dollars  per  day   (  =  ;77  )  •    The  money  earned  by 

both  men  together  is  u  -\-  v  and  their  combined  rate  of  earning  is 
5  dollars  per  day,  which  is  the  rate  of  change  of   {u  -^  v). 

32.  The  differentiation  of  the  product  of  two  functions. — Let 
u  and  V  be  any  two  functions  of  x,  and  let 

y  =  uv  (1) 

Then 

dy  =  udv  +  v-du  (2) 

Proof. — Writing  y  ■}-  Ay  for  y, 

u  -\-  Au  for  u  and 

V  -\-  Av  for  V  in  equation  (1)  we  have: 
y  +  Ay  =  {u  +  Au){v  +  Av)  (3) 

or 

y  -^  Ay  =  uv  -\-  u-Av  +  v-Au  +  Au-Av  (4) 

Whence,  subtracting  equation  (1)  from  equation  (3)  member  by 
member,  we  have 

Ay  =  u- Av  -\-  V ' Au  ■}-  Au' Av  (5) 

But    Au'Av    is  a  second  order  infinitesimal,  and  therefore  when 


FORMULAS  FOR  DIFFERENTIATION  AND  -INTEGRATION.  41 

Au  and  Av  approach  zero  we  may  drop  Au-Av  and  we  have  the 
limiting  relation: 

dy  =  U'dv  +  V'du 

Example. — Let  u  =  ax^  +  hx,  let  v  =  cx^  +  1  and  let 
y  =  {ax^  +  hx){cx^  +  !)•  To  differentiate  this  expression  substi- 
tute the  values  of  u  and  v,  and  the  values  of 

du  (=  2ax'dx  +  h-dx)     and    dv  (=  Zcx'^-dx) 
in  equation  (2),  and  we  have 

dy  =  {ax^  +  5a;)  X  ?>cx^'dx  -\-  {cx^  +  l)(2ax  -\-h)'dx 

33.  Differentiation  of  the  quotient  of  two  functions. — Let    u 

and  V  be  any  two  functions  of  Xy   and  let 

y  =  -  (1) 


Then 


^^^vdu-u-dv  (2) 


Proof. — From  equation  (1)  we  have 

.    .         u-\-  Au  .^. 

whence,  subtracting  equation  (1)  from  equation  (3)  member  by 
member,  we  have 

.         u  -\-  Au       u  ,.. 

Ay  =  — — (4) 

^       V  +  Av       V 

Reducing ; — -—  and  -  to  a  common  denominator,  equation 

V  -\-  Av  V  *     ^ 

(4)  becomes: 

V'Au  —  u-Av  ,.v 

v^  +  v-Av 


But  as   Au   and   Av   approach  zero  the  denominator  approaches 
v^    as  its  limit.    Therefore  equation  (5)  reduces  to  equation  (2). 


42  CALCULUS. 

Example. — Let  it  be  required  to  differentiate  the  function: 

Let  u  =  as?  -^rhx  and  let  v  =  cx^  -\-  d.  Substitute  these  values 
of  u  and  v,  and  the  values  of  du{=  2ax-dx  -\-h'dx)  and 
d/o{  —  Zcx^ '  dx)   in  equation  (2)  and  we  get  the  desired  result. 

34.  Differentiation  of  a  function  of  a  function. — Let    y    he  a 

Junction  of  x  and  let  z  he  a  function  of  y.    Then  y  changes  -^ 

dz 
times  as  fast  as  x,  and  z  changes  j-  times  as  fast  as  y.    There- 

dz       dv 
fore  it  is  evident  that    z    changes    t"  X  -t^   times  as  fast  as  x. 


That  is, 


^  =  -  X  ^  (I) 

dx      dy  ^  dx  ^^l 


dxi  dz 

The  symbols  -^  and  -7-    are  so  cumbersome  that  the  proposition 

upon  which  this  equation  is  based  is  not  easy  to  understand  when 
it  is  stated  as  above.  Reduced  to  its  simplest  terms  the  proposi- 
tion is  as  follows:  If  z  changes  5  times  as  fast  as  y  and  if  y 
changes  7  times  as  fast  as  x,  then  z  changes  5X7  times  as  fast 
as  x\ 

Example. — Let 

2  =  a(x2  -h  a;  -f-  l)^ 

To  differentiate  this  expression  one  can  write  y  for  a;^  +  a;  -}-  1. 

Then   z  =  ay^   and   j-  =  Zay"".    But   ^  =  2x-\-l.     Therefore, 

dz 
using  equation  (1),  we  have  t~  =  Sa{x^  +  x  -\-  l)2(2x  +  1). 

For  problems  see  group  1  in  the  Appendix. 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  43 
DIFFERENTIATION  OF  LOGARITHM. 

35.  Convergent  series. — Before  the  derivative  of  logarithm  of  x 
can  be  found  it  is  helpful  to  consider  the  two  series  of  fractions: 


a 

h 

c 

d 

e 

.  .  . 

n 

1st  series 

1 
2 

1 
4 

1 

8 

1 
16 

1 
32 

.  .  . 

1 

2» 

2d  series 

1 

II 

1 
1 

1 

II 

1 

1 

1 

where    |n   stands  for  the  product  of  all  positive  integers  from    1 
to  n  inclusive,  factorial  n,  as  it  is  sometimes  called. 

The  first  series  may  be  brought  before  the  reader  most  distinctly 
by  means  of  the  following  schedule; 

Fraction  of  original  sum 
which  is  spent  each  day. 

Half  of  a  sum  of  money  is  spent  the  first  day ^ 

Half  of  the  remainder  is  spent  the  second  day i 

Half  of  the  remainder  is  spent  the  third  day i 

Half  of  the  remainder  is  spent  the  fourth  day .  . .  tV 
etc.,  etc.,  etc. 

If  one  were  to  follow  this  schedule  indefinitely  the  original  sum 
of  money  would  never  be  entirely  spent,  because  there  is  always 
an  unspent  remainder;  but  the  remainder  would  grow  smaller 
and  smaller,  and  the  amount  spent  would  approach,  as  nearly  as 
you  please,  to  the  entire  original  sum.  This  is  equivalent  to 
saying  that  i  +  J  +  i  +  T^  +  etc.  can  never  be  equal  to  unity, 
but  it  can  be  made  to  approach  unity,  as  nearly  as  you  please,  by 
adding  together  a  larger  and  larger  number  of  the  successive 
fractions  of  the  series.  Such  a  series  of  fractions  is  called  a 
convergent  series,  and  the  limit  of  J  +  i  +  i  +  etc.,  as  a  greater 
and  greater  number  of  the  successive  fractions  are  included,  is 
called  the  sum  of  the  series.  The  sum  of  the  series  ^  +  i  +  i  + 
xV  +  etc.   is  unity. 

A  series  need  not  be  convergent  merely  because  the  successive  fractions  of 
the  series  grow  smaller  and  smaller.    Thus  |  +  i  +  i+i  +  i+etc.  grows 


44 


CALCULUS. 


a 

h 

c 

d 

1st  series 

i 

hoi  a 

iof6 

ioic 

2d  series 

J 

J  of  a 

iof6 

ioic 

larger  and  larger  without  limit  as  a  greater  and  greater  number  of  the  successive 
fractions  are  included.    Such  a  series  of  fractions  is  called  a  divergent  aeries. 

To  show  that  the  series  ,7:, ,  o  >  rj » ,-^)  etc.    is  a  convergent  series 

|2  |3   14  [5 

it  is  sufficient  to  arrange  the  two  series,  as  in  the  following  table, 
showing  how  each  fraction  is  related  to  the  preceding  fraction 
in  each  series: 


etc. 
etc. 


Each  fraction  of  the  second  series  is  equal  to  or  smaller  than 
the  corresponding  fraction  of  the  first  series.  Therefore,  if  the 
first  series  is  convergent,  the  second  series  must  also  be  convergent. 

The  particular  series  which  is  used  in  finding  the  derivative  of  a 
logarithm  is 

^=^+i+L2+ii+ii+ii+^*''-       ^'^ 

and  this  series  is  convergent  because  it  is  identical  to  the  second 
series  in  the  above  table  with  the  exception  of  the  first  two  terms. 
The  sum  of  this  series,  correct  to  seven  decimal  places,  is: 

e  =  2.7182818  (2) 

36.  The  limit  of   (  1  +  -  j   as  z  approaches  infinity. — This  limit 

is  important  in  finding  the  derivative  of  a  logarithm.     Let  us  assume 

that   z   is  always  a  positive  integer.     In  this  case  ( 1  -f  -  j 

be  expanded  by  the  binomial  theorem,*  giving: 

(■+»'-+,l'©+,i*-'>©' 


can 


+  ^z{z-l)(z-2)(^iy+  etc. 


(3) 


*  See  foot-note  to  Art.  30.    Substitute  o  =  1,  6  =  -  and  n  =  z,  and  we 

z 

have  equation  (3). 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  45 

But  as  2  approaches  infinity  (1 j,    (1 J,  etc.,  approach 

unity,  and  therefore  the  limiting  value  of  ( 1  H — )  ,  as  2  approaches 
infinity,  is  found  by  writing  unity  for  each  of  the  expressions 
(l ),  (l ),  etc.,  in  equation  (4),  and  when  this  is  done  the 

second  member  of  equation  (4)  reduces  to  1  +  il  "^  1 9  "^  I  q  "^  fl  "^ 

etc.  Therefore  the  limiting  value  of  ( 1  +  - )  as  z  approaches 
infinity  is   2.7182818,    or   e,    as  explained  in  the  previous  article. 

To  prove  that  the  limit  o/    f  1  +  -  ]     is  e  when  z  approaches  infinity,  hiU 

when  z  is  not  always  a  positive  integer.  Let  s  be  any  positive  value  whatever 
of  z,  and  let  w  be  a  positive  integer  such  that  z  lies  between  m  and  m  +  1. 

Then    (  1  +  -  )     lies  between   [  1 -\ —  )      and   {  1  +  — r— r  )      .     Now  as 
\         zj  \         mj  V      '  m  + 1/ 

(I    \m                   /  1  \m+l 

1  -\ j     and    {l-\ ^— r  1         both  approach 

e  as  a  Umit;  and  therefore     f  1  +  -  J  ,  which  is  always  between  (  1  -^ —  j 

(1      \m+i 
1  4 ^— r  j      ,   also  approaches  e  as  a  limit. 

To  show  that    (  1  H —  )      approaches    e    as  a  limit  when    z    approaches 

negative  infinity.    Let  z  =  —  r.    Then  fl-1 —  j  =fl j     where  z  is 

negative  and  r  is  positive.     Now 

but  (l+^^J  approaches  unity  as  r  approaches  infinity  and  (  H — ^  ] 


46  CALCULUS. 

approaches  e  as  r  approaches  infinity.  Therefore  every  member  of  equation 
(5)  approaches  e  as  r  approaches  infinity.  Consequently  f  1  H —  ]  ap- 
proaches e  when  z  approaches  negative  infinity. 

PROBLEM. 

One  dollar  placed  at  compound  interest  for  10  years  at  6  per 

cent,  amounts  to    (1  +  0.06)^^    dollars,  when  accrued  interest  is 

added  to  the  principal  once  a  year.     If  accrued  interest  is  added 

to  the  principle  n  times  per  year,  one  dollar  after  10  years  would 

/        0.06  \  ^^ 
amount  to     ( 1  +  — —  I     .     Find  what  one  dollar  would  amount 


('+"-?) 


to  after  10  years  at  6  per  cent,  compound  interest  (a)  when  accrued 
interest  is  added  to  principle  once  per  year,  that  is,  when  n  =  1, 
(6)  when  n  =  2,  and  (c)  when  accrued  interest  bears  interest 
without  any  delay  whatever,  that  is,  when  n  =  oo.  Ans.  (a) 
$1,791,  (6)  $1,803,  (c)  $1,822. 

Note.-  (  1  +  M§  )^"=  [(l  +\yj  where  .  =  ^  and  a  =  0.6. 

37.  Differentiation  of  the  logarithm  of  x. — Let  y  be  the  number 
of  times  that  a  given  number  a  must  be  multiplied  by  itself  to 
give  X.  Then  x  =  a^,  and  y  is  called  the  logarithm  of  x  to  the 
base  a.  The  base  of  the  common  system  of  logarithms  is  10,  and 
the  base  of  the  Napierian  system  of  logarithms  is  2.7182818  or  e, 
Napierian  logarithms  are  used  throughout  this  treatise  except  where 
it  is  otherwise  expressly  stated.* 

Let 

y  =  log  X  (1) 

Then 

dy  =  |  (2) 

Proof. — ^Writing  y  +  Ay  for  y  and  writing  x  +  Ax  for  x  in 


*  Napierian  log  x  = 


logio  X 
logxo  e 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  47 

equation  (1)  we  have: 

y  +  Ay  =  log  (x  +  Ax)  (3) 

whence,  subtracting  equation  (1)  from  equation  (3)  member  by- 
member,  we  have: 

Ay  =  log  (x  +  Ax)  —  log  X  (4) 

or 

Ay  _  log(a;  +  Ax)  -  log  x  ,  . 

(X  +  Ax\ 
J    or  to 

log  ll-j-~y,  and  ^log  (l+"^)  is  equal  to  logfl+-^j^. 

Therefore  equation  (5)  reduces  to 


£  =  '-( 


l+f)^  (6) 


and  this  may  be  written 

A^ 
Ax 


=  l,og(i+^)£  (7) 


X 

or,  writing  z  for  -— ,  we  have: 
Ax 


:=ilog(l  +  y'  (8) 

But,  as  Ax  approaches  zero   ^  {='ir-)   approaches  infinity,  and 

the  limiting  value  of    ( 1  H —  1    as  z  approaches  infinity  is  e,  as 

explained  in  Art.  36.    Therefore  as  z  approaches  infinity  equation 
(8)  becomes: 

But  the  Napierian  logarithm  of  e  is  unity  because  e  is  equal  to  eK 


48  CALCULUS. 

Therefore  equation  (9)  reduces  to 

^= -  d    =— 

dx     X  X 

Differentiation  of  loga  x. — Let 

y  =  loga  X  (10) 

This  means  that 

x  =  a«  (11) 

Therefore,  taking  Napierian  logarithm  of  each  member  of  this 
equation,  we  have 

log  X  =  2/  log  a 
or 

where  * is,  of  course,  a  constant.     Therefore,  according  to 

loga 

equation  (1)  and  (2)  we  have 

dy  =  ,-i-.^  =  l5i^l:^  (13) 

•'       log^   X  ^    X  ^ 

38.  Differentiation  of  the  exponential  function. — Let 

y  =  Ae^-  (1) 

where  A   and  k  are  constants  and  e  =  2.7182818.    Then: 

g  =  kAe^-  =  ky  (2) 

Proof. — Taking  logarithms  of  both  members  of  equation  (1) 

we  have 

log  y  =  \ogA  +  kx  (3) 

But  the  differential*  of  log  y  is  —  according  to  Art.  37.    There- 

*  It  is  very  important  that  the  student  keep  in  mind  what  a  differential  is. 
The  differential  of  log  y  is  the  infinitesimal  increment  of  log  y  due  to  an 
infinitesimal  increment,  dy,  of  y. 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  49 
fore,  by  differentiating,  both  members  of  equation  (3)  we  have: 

^  =  k-dx  (5) 

y 

dv 
whence  equation  (3)  is  obtained  by  solving  for  ^;  and  the  value  of 

y{=  Ae^")  may  be  substituted  for  y  in  the  resulting  equation. 

Differentiation  of  a**. — Let 

y  =  a*'  (6) 

where  a  and  k  are  constants.    Taking  logarithm  of  each  member 
of  (6),  we  have 

log  y  =  kx  log  a  (7) 

Differentiating  we  have 

'^^kloga-dx  (8) 

dv 
Therefore,  solving  for   -~   and  substituting  the  value  of   y  from 

equation  (6),  we  have: 

dv 

^  =  k  log  a-a**  =  k  log  a-y  (9) 

PROBLEM. 

Given  the  logarithm  of    2    to  base    10    find  the  approximate 
value  of  logarithm  of  2.01.    Ans.  0.3032. 

Note. — Consider  logio  x.    The  derivative  of  logio  x  multiplied  by  a  small 
increment  of  x  gives  the  approximate  value  of  the  increment  of  logio  x. 

39.  General  proof  of  equation  (2)  of  Art.  30. — Let 

y  =  ax"*  (1) 

where    n    has  any  value,  integral  or  fractional,  positive  or  negative.    Taking 
logarithms  of  both  members  of  (1)  we  have: 

log  y  =  log  o  +  n  log  a:  (2) 

Differentiating,  we  have 

^^  =  n^  (3) 

y  X 

5 


50 


CALCULUS. 


log  o  being  a  constant.    Substitute  ox"  for  y  in  this  equation  and  solve  for 
dy,  and  we  have 

dy  =  ruix'*~^'dx  (4) 

For  problems  see  group  2  in  the  Appendix. 

DIFFERENTIATION  OF  TRIGONOMETRIC  FUNCTIONS. 


40.  Limit  of 


sin  0 


2r  sin  p 


Fig.  18. 
chord  ob  is  equal  to  2r  sin  4>. 


as  0  approaches  zero. — It  is  a  fundamental 


principle  in  geometry  that  if  a 
polygon  of  n  sides  be  inscribed 
in  a  circle,*  the  sum  of  the 
sides  of  the  polygon  approaches 
the  length  of  the  circular  arc  as 
a  limit  when  n  approaches  in- 
finity. The  idea  that  a  circu- 
lar arc  has  a  definite  length  de- 
pends upon  this  principle. 

Consider  Fig.  18.    The  arc 
ah   is  equal  to  2r0,   <f)  being 
expressed  in  radians,  and  the 
Therefore : 


chord  ah      2r  sin  <^       sin  <f> 


arc  ah 


2r0 


But  approaches  imity  as  its  limit  when    0    approaches 


arc 


zero;  and  therefore 


sm  (f> 


must  also  approach  unity  as  its  limit 


when  (f)  approaches  zero. 
41.  Differentiation  of  sin  x. — Let 

y  =  sin  X  (1) 

dy  =  cos  x-dx  (2) 

Proof. — Writing  y  -\-  Ay  for  y  and  writing  x  +  Ax  for  x,  we 


Then 


*  Or  any  continuous  curve. 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  51 

have : 

y  -{•  Ay  =  sin(x  +  Ax)  (3) 

Whence,  subtracting  equation  (1)  from  equation  (3)  member  by- 
member  and  solving  for   -^ ,  we  have 
Ax 

Ay  _  sm(x  +  Ax)  —  sin  x 
Ax  ~  Ax 


(4) 


Using  the  trigonometric  formula: 

sin  a  —  sin  6  =  2  sin  ^{a  —  h)  cos  ^{a  +  6)  (5) 

equation  (3)  may  be  reduced  to: 


Ay  _  sin  jAx 


•cos(x  +  J  Ax)  (6) 


Ax  |Ax 

but  as  Ax  approaches  zero  cos(x  +  J  Arc)  approaches  cos  x  as 
its  limit,  and  -^~ —  approaches  unity  as  its  limit.  Therefore, 
from  equation  (6)  we  have : 

-^  =  cosx    or    dy  =  cosx-dx 

42.  Differentiation  of  cos  x. — Let 

y  =  cos  X  (1) 

Then 

dy  =  —  sinx-dx  (2) 

Proof. — Proceeding  as  in  the  previous  article  we  get 

Ay  _  cos(a;  +  Ax)  —  cos  x 
Ax  Ax 


(3) 


and  using  the  trigonometric  formula: 

cos  a  —  cos  h  —  —  2  sin  i(a  —  6)sin  i(a  +  h)  (4) 

we  obtain 

Ay  sinjArc    .   /     ,    i.   n  /kx 


52  CALCULUS, 

and  from  this  we  obtain 

-^  =  —  emx    or    dy  —  —  Qmx'dx 

Far  problems  see  group  3  in  the  Appendix. 

EXAMPLES  SHOWING  USE  OF  FORMULAS  FOR 
DIFFERENTIATION. 

43.  Differentiation  of    tan  x. — In  order  to  differentiate   tan  x 

sin  X 

the  familiar  formula  tan  x  = may  be  used  as  follows: 

cos  a; 

,  sinx  ,^. 

y  =  tan  x  = (1) 

cos  a; 

Proceeding  according  to  Art.  33,  let  w  =  sin  x  so  that  du  = 
cos  x  •  dx,  and  let  v  =  cos  x  so  that  dv  =  —  sin  x  •  dx.  Then 
using  equation  (2)  of  Art.  33,  we  have: 

J        cos^  X'dx  +  sin^  x-dx        dx  /„^ 

dy  =  ^ =  — ^  (2) 

COS^  X  COS^  X 

Any  trigonometric  function  can  he  differentiated  hy  expressing  the 
function  in  terms  of  sine  and  cosine  and  using  the  formulas  of  Arts, 
41  and  4^. 
As  a  further  example,  consider 

i  =  I  sin  (at  (3) 

in  which  I  and  to  are  constants  and  t  is  clasped  time  reckoned 

from  any  chosen  instant,  and  let  it  be  required  to  find  the  rate  of 

di 
change  of  i,  namely,   -jz.    Now  cot  is  a  function  of  t  and  sin  <at 

is  a  function  of  (at.  Therefore  we  may  use  the  formula  for  the 
differentiation  of  a  function  of  a  function,  as  explained  in  Art.  34. 
Let 

z  =  (at  (4) 

then 

dz  =  (a-dt  (5) 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  53 

Writing  z  for  wt  in  equation  (3)  we  have : 

i  =  7  sin  2  (6) 

which,  by  differentiation  gives: 

di  =  I  cos  Z'dz  (7) 

Whence,  substituting  ot  for  z  and  03 -dt  for  dz  we  have 

di  =  0)1  cos  cot'dt  (8) 

or 

-r:  =  co7  cos  co<  (9) 

at 

For  problems  see  group  4  ^^  t^^  Appendix. 

44.  Differentiation  of  sin*^  x. — The  inverse  trigonometric  func- 
tions can  all  be  differentiated  with  the  help  of  the  formulas  al- 
ready established.     For  example  let 

y  =  sin-^  X  (1) 

this  means  that 

siny  =  X  (2) 

which,  by  differentiation,  gives: 

cos  y-dy  =  dx  (3) 

but  cos 2/  =  Vl  —  sm^y  =  V 1  —  x^,  according  to  equation  (2). 
Therefore,  substituting  this  value  for  cos  y  in  equation  (3)  and 
solving  for  dy,  we  have 

For  problems  see  group  6  in  the  Appendix. 

45.  Differentiation  of  u^. — Let  u  and  v  be  any  two  functions 
of    Xf    and  let 

y  =  u-  (1) 

To  differentiate  this  expression  take  logarithms  of  both  members 
and  we  have 

\ogy  -=  V  log  u  (2) 


54  CALCULUS. 

This  is  a  product  of  two  functions  v  and  logw  and  their 
respective  differentials  are  dv  and  — .  Therefore,  using  the 
method  of  Art.  32,  we  may  differentiate  equation  (2),  and  we  have: 

dy      V'du  ,  I  ,  /ON 

—  = +  logU'dv  (3) 

y         u  '=' 

whence,  substituting  u"  for  y  and  solving  for  dy  we  have 

dy  =  vu*'~^du  +  log  uu^-dv  (4) 

When  u   and   v   are  known  as  functions  of  x,   then  du  and  dv 
are  known  in  terms  of  x  and  dx,  and  the  known  values  of  w,  r, 
du  and  dv  may  be  substituted  in  equation  (4)  thus  giving  dy  in 
terms  of  x  and  dx. 
For  problems  see  group  6  in  the  Appendix. 

SUCCESSIVE  DIFFERENTIATION  AND  INTEGRATION. 

46.  Successive  derivatives  of  a  function. — Consider  a  function 
of  x.    For  example: 

y  =  ax^  (a) 

then 

I  =  '^  (« 

But  this  derivative  is  itself  a  function  of   x   and  its  derivative  is 

12ax^;    this  is  also  a  function  of    x    and  its  derivative  is    24aa;; 

and  so  on.    These  successive  derivatives  of  the  original  function 

du    CbU    d  u 
y  are  usually  represented  by  the  symbols  —,  -~,  -^  and  so  on; 

so  that  we  may  write 

dx^ 


Uax"  (2)^ 


g  =  24ax  (3) 

etc.        etc. 


*kM}' 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  55 

Equations  (1),  (2),  (3),  etc.,  express  what  are  called  the  first 

derivative^  the  second  derivative,  the  third  derivative,  etc.,  of  y. 

The  exact  meaning  of  the  successive  derivatives  of    y   may  be 

most  easily  expressed  in  words  when   x   represents  elapsed  time. 

du  d^ij 

Then  ~    is  the  rate  of  change  of  y,   -t\  is  the  rate  of  change  of 

dv     d^y  d^ij 

^,    -7-^  is  the  rate  of  change  of  ~,  and  so  on.     In  the  particular 

example,  namely    y  =  ax^,    the  fourth  derivative  is  a  constant 
(=  24a),  and  the  fifth  and  all  higher  derivatives  are  equal  to  zero. 

47.  Velocity  and  acceleration.  Use  of  first  and  second  deriva- 
tives.— In  the  formulation  of  problems  in  mechanics  one  must 
frequently  express  the  velocity  and  acceleration  of  a  moving  body 
in  terms  of  its  coordinates.  Thus  x  and  y  are  the  varying 
coordinates  of  a  moving  particle  B  in  Fig.  19;  x  and  y  have 
definite  values  at  each  instant  as  the  body  moves,  and  they  are 
therefore  functions  of  elapsed  time;  hence  we  have  the  following 
important  relations: 

dx 

-^  is  the  x-component  of  the  velocity  of  B. 

dv 

-jj   is  the  ^/-component  of  the  velocity  of  B. 

-77^  is  the  a;-component  of  the  acceleration  of  B, 

-^   is  the  ^/-component  of  the  acceleration  of  B, 

These  relations  are  evident  from  the  following  considerations: 

Let  Ax  and  Ay  be  the  increments  of  x  and  y,  respectively,  during 

Ax 
a  short  interval  of  time.    Then  —   is  the  average  velocity  of  the 

body  in  the  direction  of  the  a:-axis  during  the  interval    At,    and 

dx  Ax 

-r.   is  the  limiting  value  of   —  when  A^  and  Ax  approach  zero. 

Av 
Similarly   -^   is  the  average  velocity  of  the  body  in  the  direction 
At 


56 


CALCULUS. 


of  the  j/-axis,  and  -^  is  the  limiting  value  of  -^  when  At  and  Ay 
approach  zero.    Furthermore  the  x-component  of  the  acceleration 


iHixtti; 

baseball 


B 


x-axU 


X 

Fig.  19. 


Fig.  20. 


of    B    is  equal  to  the  rate  of  change  of  the  a;-component  of  the 

velocity  of  5,   that  is,  the  a;-component  of  the  acceleration  of  B 

dx  d/^x 

is  the  rate  of  change  of  -^r   which  is  -^. 

Example. — ^A  ball  travels  at  uniform  angular  velocity  (w  radians 
per  second)  around  a  circle  of  radius  A  as  shown  in  Fig.  20. 
In  this  case  the  coordinates  x  and  y  are  simple  functions  of  the 
elapsed  time  t;  indeed  from  Fig.  20  we  have: 


a;  =  A  cos  wt 
y  =^  A  sin  <t)t 


Differentiating  these  expressions  with  respect  to  <  we  have: 

dx 
and 


dt 


=  —  w  A  sin  ui 


-^  =  (aA  cos  (i)t 


(1) 

(2) 


(3) 


(4) 


These  equations  express  the  x  and  y  components  of  the  velocity 
of  the  ball    B   at  each  instant.    Differentiating  (3)  and  (4)  with 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  57 
respect  to  <  we  have: 

(5) 

(6) 


-7i5    =    —  03^ A  COS  (d 


-^  =  —  o)^A  sin  (at 
ar 


These  equations  express  the  x  and  y  components  of  the  acceler- 
ation of  the  ball  B  at  each  instant. 


PROBLEMS 


1 .  If  X  =  tan  6  -f  sec  0,    prove  that 


cos  0 


cPx 


dff"     (1  -  sin  ey 


2.  If  2/  =  x^  log  X,  prove  that   j:J  =  ~« 

3.  If  1/  =  e-loga;,  prove  that  g  =  e*^^og  «+~|+|-|)- 

4.  If  2/  =  c~*  cos  a;,  prove  that  -7^  +  ^2/  =  0. 

d^x 

5.  If  a;  =  A  sin  n<  +  B  cos  n<,   prove  that  -77^  +  n^a;  =  0. 

6.  For  what  values  of  a  will  y  =  e*^  satisfy  the  equation 

da^      da;^         da; 
Ans.    0,  -  1,  +  2. 

7.  A  ball  moves  so  that  its  x  and  y  coordinates  are  At  cos  d 
and  At  sin  0,  respectively,  as  shown  in  Fig.  p7,  where  A  and  0 
are  constants  and   t   is  elapsed  time.    Find  (a)  the  a;-component 


58  CALCULUS. 

and  (6)  the  y-component  of  the  velocity  of  the  ball,  and  (c)  the 
a;-component  and  (d)  the  y-component  of  its  acceleration. 
Ans.     (a)   A  cos  d,    (b)   A  sin  d,   (c)   0,    (d)   0 
8.  A  ball  moves  so  that  its  x  and  y  coordinates  are  A^  cos  ^ 
and    Bt^  sin  6,    respectively,  where   A,    B    and    0   are  constants 
and    t    is  elapsed  time.     Find  (a)  the  x-component  and  (6)  the 
^/-component  of  the  velocity  of  the  ball,  and  (c)  the  x-component 
and  (d)  the  ^/-component  of  its  acceleration. 
Ans.     (a)   A  cos  6,   (h)   2Bt  sin  d,    (c)   0,    (d)   2B  sin  6. 


9.  The  resultant  acceleration  of  a  body  is 


mHM' 


dt^ 
as  shown  in  Fig.  p9,  where  </>   is  the  angle  whose  tangent  is  ^-  * 

dt^ 
Show  from  equations  (1)  to  (6)  of  Art.  47  that  the  resultant 
acceleration  (or  total  acceleration)  of  the  ball    B    in  Fig.  20  is 

equal  to  -j ,  and  show  that  it  is  parallel  to  the  radius  A  at  each 

instant,  v  being  the  resultant  velocity  \/  (  ;^  )   +  (  7^  )  • 

10.  If  the  distance  traveled  by  a  body  is  s  =  ae*'  +  6e~**, 
show  that  the  acceleration  is  a^s.  If  t  is  expressed  in  seconds 
and  s  in  feet,  (a)  in  what  units  is  a  expressed,  and  (6)  in  what 
units  is  ah  expressed? 

Ans.     (a)  reciprocal  seconds,    (6)  feet  per  second  per  second. 

48.  Harmonic  motion.  Example  showing  the  use  of  a  second 
derivative. — A  body  m  is  suspended  by  a  spring  and  the  body 
stands  in  equilibrium  in  the  position  shown  in  Fig.  21.  If  the 
body  is  displaced  x  feet  downwards  (or  upwards)  an  unbalanced 
force  F  will  act  upon  the  body,  and  this  unbalanced  force  will 
be  proportional  to  x.    Therefore  we  may  write 

F  =  -kx  (1) 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION- 59 


The  negative  sign  is  chosen  because    F    is  upwards  when    x    is 

downwards,  and  F  is  downwards  when  x   is  upwards. 

If  the  body  is  pulled  upwards  or  downwards  and 

released,  it  will  vibrate  up  and  down,  and  x    will 

dx 
vary  with  the  time    t.    Then  -rr  is  the  velocity 


dt 

d^x 
of  the  body  at  each  instant,  and  -^    (the  rate  of 

dx\ 
change  of   -77  )  is  the  acceleration  of  the  body  at 

each  instant.     But  the  unbalanced   force  which 
acts  upon  a  body  is  equal*  to  mass  X  acceleration. 

Therefore  F  =  m  >  -p,  or,  substituting  the  value 

of  F  from  equation  (1),  we  have: 


spring 


df 


m 


(2) 


To  determine  the  motion  of  the  body  in  Fig.  21 
it  is  necessary  to  find  a;  as  a  function  of  t  which 
will  satisfy  the  law  of  growth  as  expressed  by  equa- 
tion (2),  and  the  problem  is  solved  if  we  find  a 


k 


function  of  t  whose  second  derivative  is  equal  to multiplied 

by  the  function  itself.  This  function  would  be  at  once  recognized 
if  one  were  familiar  with  the  derivatives  of  all  kinds  of  functions. 
Thus  we  have  another  example  showing  how  important  it  is  to 
be  familiar  with  the  derivatives  of  functions.  Compare  Arts. 
22  and  23. 

Consider  the  function 

a;  =  a  sin  (coi  +  6)  (3) 

where  a,  w  and  0  are  undetermined  constants.     Differentiatingf 
*  When  mass  is  expressed  in  pounds,  acceleration  in  feet  per  second  per 

second,  and  force  in  poundals;  or  when  C.G.S.  units  are  used  throughout, 
t  Let  z  =  Oil  ■\-  B  and  use  the  formula  for  the  differentiation  of  a  function 

of  a  function  as  explained  in  Art.  34. 


60  CALCULUS, 

equation  (3)  with  respect  to  t  we  have: 

^  =  coa  cos  (o)t  -f  0) 

and  differentiating  again,  we  have: 


'«! 


cPx 

^  =  -  co^a  sin  (ut  +  6)  (6) 


or,  substituting  x  for  a  sin  {cot  -\-  0),  we  have: 

<Px 


^,  =  -<^  (6) 


Therefore,  if 

„       k 


m 


(7) 


equation  (3)  gives  the  required  function,  and  a  and   B  can  have 

any  values  whatever. 

If  the  vibrations  are  started  by  drawing  the  body  down  to 

a?  =  A,    and  if  time  is  reckoned  from  the  instant  of  release,  then 

dx 
X  —  A  when  <  =  0,   and  also  -r:  =  0  when  f  =  0   because  the 

at 

body  has  no  velocity  at  the  instant  of  release.    Using  the  values 
:^  =  0    and    «  =  0    : 
equation  (3)  becomes; 


dx 

-ji  —  ^    and    i  =  0    in  equation  (4),  we  find    0  =  0.    Therefore 


aj  =  a  sin  w<  (8) 

Using  the  values  x  =  A  and  f  =  0  in  equation  (8),  we  find 
a  —  A»  Therefore,  jor  the  case  in  which  the  body  is  pulled  down  to 
X  =  A  and  released  when  t  =  0,  equation  (8)  becomes: 

a;  =  A  sin  cot  (9) 

The  motion  which  is  defined  by  equation  (9),  or  in  general  by 
equation  (3),  is  called  simple  harmonic  motion.    The  distance    x 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  61 


Fig.  22. 


of  the  vibrating  body  from  its  equilibrium  position  is  at  each 
instant  proportional  to  the  sine  of  a 
uniformly  growing  angle  o)t;  or,  the 
position  P  of  the  vibrating  body  is 
at  each  instant  the  projection  on  the 
straight  line  ah  of  a  point  P'  which 
moves  round  the  circle  CC  at  con- 
stant angular  velocity  w,  the  radius 
of  the  circle  being  A,  as  shown  in 
Fig.  22. 

One  complete  vibration  of  the  body  in 
Fig.  21  takes  place  while  the  point  P'  in 
Fig.  22  makes  one  complete  revolution, 
that  is,  while  the  angle  (at  increases  from  zero  to  2t  radians,or  while 

the  time  t  increases  from  zero  to  —  seconds,  to  being  expressed 

CO 

in  radians  per  second.    Therefore  — ,  or  27r  -^  a/—  [see equation 

(7)],  is  the  period  of  one  complete  vibration.    That  is  27r  -^  -W— 

fk 
is  the  number  of  seconds  per  vibration,  or  -yl—   4-  2ir  is  the  number 

of  vibrations  per  second,  or  frequency,  of  the  vibrating  body. 

In  this  discussion  the  mass  of  the  spring  and  the  effects  of  friction 
are  ignored. 

49.  Curvature. — Consider  the  curve  cc  in  Fig.  23.  The 
tangent  line  a  touches  the  curve  at  p  and  the  tangent  line  h 
touches  the  curve  at  q.  Therefore  the  tangent  line  turns  through 
the  angle  ^  when  the  point  of  tangency  moves  from  p  to  q. 
The  sharper  the  bend  or  curvature  of  the  line  cc  the  greater  the 
value  of  j8  for  a  given  length  of  arc  pq.  Dividing  the  angle 
iS  (expressed  in  radians)  by  the  length  of  the  arc  pq  we  have  a 
quotient  which  is  used  as  a  measure  of  the  average  curvature  of  cc 
between  p   and   q,   and  the  limiting  value  of  this  quotient  when 


62 


CALCULUS, 


the  arc   pq  approaches  zero  is  the  measure  of  the  curvature  of  cc 
at  the  point  p.* 

It  is  required  to  find  an  expression  for  the  limiting  value  of  the 

quotient  — - —    in  Figs.  23  and  24  when  the  arc  pq  approaches 
arc  pq 

zero.    Let  the  coordinates  of  p  be  x  and  y,  and  let  the  coordi- 


Fig.  23.  Fig.  24. 

nates  of  q  be  x  -\-  Ax  and  y  +  Ay  as  shown  in  Fig.  24.    Then: 

chord  pq  =  V(Ax)2  +  {AyY  (1) 

and  when  the  arc  pq  approaches  zero  the  arc  becomes  more  and 
more  nearly  equal  to  the  chord  so  that  the  limiting  form  of  equation 
(1)  is: 

arc  pq  =  V(dx)2  +  {dy^  (2) 

Furthermore  we  have: 


X  dy 

tan  a  =  -5^ 

ax 


(3) 


according  to  Art.  18,  where  a   is  the  angle  shown  in  Fig.  23;  and 

dv 
of  course  tan  a'   (see  Fig.  23)  is  the  value  of  ~   at  the  point  q. 


But  ^  changes  ^  times  as  fast  as  x,  and  therefore  the  change 


*  Or  at  q,  because  p  and  q  approach  coincidence  as  the  arc  approaches 
zero. 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  63 

dv  .    d^v 

in  the  value  of  —  from  p  to  q  is  ~'  ^^»  when  Ax  is  infinitely 

small.     Therefore  the  value  of   -r    at    q   is    -~  +  -^K  '  dx,    and 

ax  dx      dx^ 

consequently  we  have: 

Now  the  angle  jS  in  Fig.  23  is  equal  to  a!  —  a,  and,  since  a!  —  a 
is  an  infinitesimal  angle,  we  may  write : 

tan  {a'  —a)  =  a'  —  a 


Therefore  we  have: 

^  =  a'  —  a  ^  tan  {a!  —  a) 
But 

(5) 

.       ,  ,         s         tan  a'  —  tan  a 

tan  (a'  —  a)  =  .,    ,    . ^-r 

1  +  tan  a'  tan  a 

(6) 

Therefore,  substituting  the  values  of  tan  a  and  tan  a!  from 
equations  (3)  and  (4),  and  writing  j8  for  tan  {a!  —  a)  according 
to  equation  (5),  we  have: 

^•dx 
dx^ 

^~^^x\dx'^d^^'^y 
But  the  infinitesimal  term  in  the  denominator  (  -~  •  dx  J  may  be 

discarded  because  it  is  added  to  a  finite  quantity  (  -r  j-     Therefore 

equation  (7)  becomes: 

^-^  .  dx 
dx^ 

P  =  /...x.  (8) 


1  + 


\dx) 


64  CALCULUS. 

Therefore,  using  the  value  of   arc  pq   from  equation  (2)  we  have 


fi 


dx* 


dx 


arc  pq 


'^(si 


^(dxy  +  {dyy 


(9) 


This  is  the  required  limiting  value  of 


i8 


because  equations  (2) 


arc  pq 

and  (5)  are  limiting  forms  and  because  an  infinitesimal  has  been 

discarded  in  going  from  equation  (7)  to  equation  (8).     Dividing 

dx 

by  dx  this  factor 


numerator  and  denominator  of 
1 


becomes 


Mty 


^{dxY  +(d2/)2 
and  equation  (9)  becomes: 


Curvature  at  p  = 


dx" 


[■  -  (2)T 


(10) 


Curvature  of  a  circle. — The  idea  of  curvature  is  necessarily 
vague  until  we  imderstand  clearly  the  unit  in  terms  of  which 

curvature  is  expressed.  Therefore 
let  us  apply  the  above  definition 
of  curvature  to  a  circle.  The 
line  a,  Fig.  25,  is  tangent  to  the 
circle  at  p  and  h  is  tangent  to  the 
^     circle  at  5,  and  the  measure  of  the 

curvature  of  the  circle  is  — - — -, 
arc  pq 

But  the  angle   j8,    between  the 

tangent    lines   is    equal    to   the 

angle  between  the  radii   Cp   and 

Cq,    Therefore  the  angle  i3,  in  radians,  is  equal  to  arc  pq  divided 

by  the  radius    r    of  the  circle,  and  consequently  the  quotient 


Fig.  25. 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  65 

— is  equal  to   -.     That  is,  the  curvature  of  a  circle  is  meas- 

arc  pq  r 

ured  by  the  reciprocal  of  its  radius,  or  the  reciprocal  of  the  curva- 
ture of  a  circle  is  equal  to  the  radius  of  the  circle.  Therefore  the 
reciprocal  of  the  curvature  of  any  curve  at  a  point  is  called  the  radius 
of  curvature  of  the  curve  at  the  point.  In  accordance  with  this 
statement  equation  (10)  may  be  written: 

[i  +  (^YV 

{radius  of  curvature  of  any  "I  _  L         \dx/  J  .^ .  ^ 

curve  at  a  point  J  d^y 

1^ 

the  coordinates  of  the  point  p  being  substituted  in  the  general 

£      dy        ,  d-y 
expressions  for  -^   and  -r%. 

Remark. — At  a  point  where  a  curve  is  horizontal  (parallel  to 

du 
the  X-axis)  the  first  derivative   -p  is  equal  to  zero,  and  at  such  a 

point  the  expression  for  radius  of  curvature  reduces  to  -j^.    Where 

a  curve  is  very  nearly  horizontal  the  first  derivative   -^   is  small 

as  compared  with  unity,  the  expression    1  +  ( ;p  )    is  sensibly 

equal  to  unity,  and  the  radius  of  curvature  is  sensibly  equal  to 
1 


dx' 

Example. — Consider  the  parabola  whose  equation  is    y  =  x^, 

dv  d^xi 

Then  -^  =  2x  and  ■—  =  2.    To  find  the  radius  of  curvature  at 

the  origin  of  coordinates,  substitute  the  values  -p  =  2x  =  0  and 
6 


66  CALCULUS. 

~  =  2  in  equation  (11)  which  gives  R=  2.  To  find  the  radius 
of  curvature  at  the  point  x  =  1  and  2/  =  1,  substitute  the  values 
_|  =  2x  =  2  and^J  =  2  in  equation  (10)  which  gives  R  =  5.6.* 

Further  discussion  of  curvatiwe.    The  osculating  circle. — A 

straight  Une  may  coincide  with  a  curve  at  a  point  and  have  the  same 

direction  as  the  curve  at  the  point.     Such  a  straight  line  is  called 

a  tangent  line. 

A  circle  may  coincide  with  any  curve  at  a  point,  have  the  same 

direction  as  the  curve  at  the  point,  and  have  the  same  curvature  as 

the  curve  at  the  point.     To  coincide  with  the  curve  at  a  point 

means  that  the  ordinate  y  of  the  circle  is  equal  to  the  ordinate  y 

of  the  curve  for  a  certain  abscissa  x.     To  have  the  same  direction 

dy  dy 

as  the  curve  at  the  point  means  that  -^  for  the  circle  is  equal  to  -~ 

for  the  curve  for  the  same  value  of  x.    To  have  the  same  curva- 

ture  as  the  curve  at  the  point  means  that  -v^  for  the  circle  is 

equal  to  —^  for  the  curve  (if  the  condition  of  tangency  is  satisfied) 

for  the  same  value  of    x.    The  circle  which  satisfies  these  three 

*  Whenever  an  equation  between  x  and  y  is  to  be  plotted  as  a  curve^ 
and  whenever  definite  values  are  to  be  determined  for  the  derivatives  of  y 
with  respect  to  x,  the  question  arises  as  to  what  units  are  to  be  used  in  the 
expression  of  y  and  its  various  derivatives.  Quantities  such  as  2  inches, 
5  pounds  and  6  hours,  are  called  denominate  quantities.  The  ratio  of  two 
denominate  quantities  of  the  same  kind  is  called  a  pure  number.  Both  mem- 
bers of  an  equation  and  every  separate  term  in  each  member  must  have  the 
same  denomination.  Consider  the  equation  y  =  ax^.  In  order  to  plot  this 
equation  as  a  curve,  x  and  y  must  both  be  expressed  in  inches,  let  us  say. 
Therefore  the  coefficient  o  is  a  denominate  quantity  which  gives  inches  when 
it  is  multiplied  by  inches  squared.  That  is,  the  reciprocal  of  a  is  expressed  in 
inches.  When  an  equation  such  as  ?/  =  4x'  is  to  be  plotted  as  a  curve,  x 
and  y  must  both  be  expressed  in  inches,  and  the  coefficient  4  must  be  thought 
of  as  a  denominate  number,  the  denomination,  or  unit,  being  such  as  will  give 
y  in  inches  when  x  is  expressed  in  inches. 


Hi 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  67 

conditions  at  a  point  on  a  curve  is  called  the  osculating  circle  at 
the  point,  and  the  radius  of  the  osculating  circle  is  called  the 
radius  of  curvature  of  the  curve  at  the  point. 

Determination  of  the  osculating  circle. — One  can  easily  find 

the  values  of  -^  and  J  at  a  given  point  p  on  a  given  curve  cc 

(see  Fig.  26)  by  differentiating  the  equation  of  the  curve. 
Consider  any  circle  whatever  as  shown  in  Fig.  27.    The  equation 


y-axi8 


( X >i 

Fig.  26. 
of  this  circle  is 


■a  — 
~x- 


Fig.  27. 


{x  -  ay  +  (2/  -  hf 


(12) 


where  a  and  h  are  the  coordinates  of  the  center  of  the  circle 
and  r  is  the  radius  of  the  circle.  Differentiating  equation  (12), 
we  have 

2{x  -  a)  '  dx  +  2{y  -h)  '  dy  =  0 
or 

{x-a)  +  {y-  h)u  =  0  (13) 

du 
where    u    is  written  for    -~.    Differentiating  equation  (13),  re- 
membering that  y  and  u  are  both  functions  of  x,  we  have: 
dx  -\-  {y  —  h)  •  du  -{-  u  *  dy  =  Q 


68  CALCULUS. 

Dividing  through  by  dx  and  remembering  that  w  •  ^  =  w^  we 
have: 

1  +  (2/  -  6)  •  g  +  w^  =  0 

du  (Pv 

But    ^    is  the  second  derivative    ~.    Therefore,  representing 

^  by  V,  we  have: 

l-\-{y-h)v-\-u'  =  0 
or 

2/  -  6  = -—  (14) 

and,  substituting  this  value  of  {y  —  h)  in  equation  (13),  we  have: 

V 

Substituting  the  values  of  (x  —  a)  and  {y  —  h)  from  equations 
(15)  and  (14)  in  equation  (12),  we  get: 

,  =  a±j^'  (16) 

Now  u  and  v  and  x  and  y  are  all  known  at  a  given  point  p 
of  the  curve  cc  of  Fig.  26,  and  these  four  quantities  u,  v,  x  and  y 
have  the  same  values  for  the  osculating  circle  where  it  touches  the 
curve  cc  at  the  point  p.  Therefore  a  and  h  can  be  found 
from  equations  (14)  and  (15),  and  r  can  be  determined  from 
equation  (16)  so  that  the  osculating  circle  at  the  point  p  is  com- 
pletely determined. 

PROBLEMS. 

1.  Find  the  radius  of  curvature  at  the  point  p  of  the  parabola 
shown  in  Fig.  pi. 
Ans.    40.5  inches. 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  69 


2.  Find  the  radius  of  curvature  of  the  ellipse  shown  in  Fig.  p2, 
(a)  at  the  point  p,  and  (6)  at  the  point  p'. 


Fig.  pi. 

Ans.     (a)  2.67  inches,  (b)  18  inches. 

dy  . 


Fig.  p2. 


Note. — The  value  of  -^  is  infinite  where  a  curve  is  parallel  to  the  i/-axis 

and  generally  the  value  of  ^  is  also  infinite  at  such  a  point.    Therefore  the 

expression  for  radius  of  curvature  at  such  a  point  of  a  curve  usually  becomes 

indeterminate.     It  is  evident,  however,  that  x  and  y  may  be  interchanged 

throughout  the  entire  discussion  of  Art  49.    This  interchange  gives  an  ex- 

dx  d^x 

pression  for  radius  of  curvature  in  terms  of  t-  and  ^  where  x  is  thought 

of  as  a  function  of  y]  and  this  expression  can  be  used  to  determine  the  radius 
of  curvature  at  a  point  on  the  curve  where  the  curve  is  parallel  to  the  y-axis. 


Fig.  p3. 


18  inches 

Fig.  p5. 


3.  A  stone  arch  is  to  be  made  as  shown  in  Fig.  p3,  the  two 
dotted  circles  having  the  same  curvature  as  the  ellipse  at  the 
junction  points  pp.    Find  the  radius  of  the  circles. 

Ans.    14.07  feet. 


70 


CALCULUS. 


center  of  curve 


4.  Find  the  radius  of  curvature  of  the  curve   y  =  sinx   at  the 
points  (a)   a;  =  0  and  (h)   a:  =  ^. 

Ans.     (a)  infinity,  (6)  2.6. 

5.  Find  the  radius  of  curvature  at  the  point    p    of  the  sine 

curve  shown  in  Fig.  p5. 

Ans.     5.47  inches. 

6.  A  straight  portion  of 
railway  track  merges  gradually 
into  a  circular  curve  as  indi- 
cated by  the  dotted  line  in 
Fig.  p6.  The  dotted  portion 
is  called  a  transition  curve. 
The   curvature  of    the   track 

(which  is  sensibly  equal  to  ^~ 


RM. 


100  feet 

Fig.  pQ. 


because 


dy     .  , 

~    IS   very   nearly 


zero)  increases  uniformly  from  zero  at  x  =  0  to  t^ftf:  at  a:  =  100. 

Find  the  equation  of  the  dotted  transition  curve. 
Ans.    900,000  y  =  a^. 

Note. — If  the  curvature  v  2  increases  uniformly  its  rate  of  change,  ^-j, 
is  a  constant. 

50.  Geometric  differentiation.  The  acceleration  of  a  particle 
which  travels  in  a  circular  orbit. — There  are  cases  in  which  the 
instantaneous  rate  of  change  of  a  given  quantity  may  be  found 
when  the  quantity  is  specified  in  geometric  terms.  An  important 
example  is  the  following:  A  particle  travels  round  and  round  a 
circular  path  or  orbit  at  velocity    v,    and  it  is  required  to  find  the 

deceleration  ^  of  the  particle. 

At  a  certain  instant  the  particle  is  at  P,  Fig.  28,  and  its  velocity 


FORMULAS  FOR  DIFFERENTIATION  AND   INTEGRATION.  71 

at  this  instant  is  represented  in  direction  and  magnitude  by  the 
Hne    vi.    During  a  short  interval  of  time  the  particle  will  travel 


Fig.  28. 


Fig.  29. 


the  distance  v.dt,  the  particle  will  then  be  at  Q,  and  its  velocity 
will  be  as  represented  by  the  line  v^  which  has  the  same  length  as  Vi. 

From  any  fixed  point  0'  draw  two  lines  parallel  and  equal  to 
vi  and  V2  respectively,  as  shown  in  Fig.  29.  Then  the  line 
dv  represents  the  velocity  which  must  be  added*  to  Vi  to  give  i;2, 
that  is  the  line  dv  represents  the  increment  of  the  velocity  of  the 
particle  during  the  interval  dt. 

Now  when  dt  is  chosen  smaller  and  smaller,  the  angle  <(>  ap- 
proaches zero,  and  the  angles  at  P'  and  Q'  approach  90°  so  that 
dv  becomes  more  and  more  nearly  parallel  to  the  radius  OP. 
Also  when  dt  is  very  small  the  arc  PQ  ( =  v-dt)  is  sensibly  a 
straight  line  and  the  two  triangles  OPQ  and  O'P'Q'  are  similar. 
Therefore  from  these  similar  triangles  we  have : 


r  :  v.dt  :  :v:dv 


(1) 


In  this  proportion  v  is  written  instead  of  Vi  or  V2,  both  of  which 

*  We  are  here  concerned  with  what  is  called  vector  addition  or  geometric 
addition.  The  most  familiar  example  of  this  kind  of  addition  is  the  addition 
of  two  forces  by  the  principle  of  the  "parallelogram  of  forces." 


72 


CALCULUS. 


have  the  same  numerical  value.    Solving  this  proportion  we  have; 


dv 
dt 


(2) 


We  have  already  proceeded  to  the  limit  in  considering  the  arc 
PQ  a  straight  line.  That  is,  the  acceleration  of  a  particle  moving 
in  a  circular  orbit  is  equal  to  the  square  of  the  velocity  of  the 
particle  divided  by  the  radius  of  the  orbit,  and  the  acceleration  is 
at  each  instant  towards  the  center  of  the  circle  because  the  infin- 
itesimal increment  of  velocity  dv  in  Fig.  29  is  in  the  direction  of 
PO  in  Fig.  28. 

51.  Geometric  differentiation.  The  inward  force  per  unit 
length  of  a  barrel  hoop. — Another  important  case  of  geometric 
differentiation  is  the  following:    Figure  30  represents  a  hoop 

around  a  tank.    It  is  required  to  find  the  value  of   -r     where    dF 

is  the  force  with  which  a  short  length  ds  of  the  hoop  pushes  inwards 
on  the  tank. 

The  radius  r  of  the  tank  and  the  tension  T  of  the  hoop  are 
given.  The  tension  of  the  hoop  is  the  force  with  which  a  portion 
of  the  hoop  pulls  on  an  adjoining  portion.    Thus  the  two  forces 


tank 


Fig.  30. 


Fig.  31. 


FORMULAS  FOR  DIFFERENTIATION  AND  INTEGRATION.  73 

Ti  and  T2,  Fig.  30,  which  pull  on  the  portion  ds  of  the  hoop 
in  Fig.  30,  are  each  equal  to  the  tension  T,  and  the  resultant 
of  the  two  forces  Ti  and  T2  is  the  total  inward  force  dF  exerted 
on  the  portion;  and  this  total  inward  force  acting  on  ds  is  the 
force  with  which  ds  pushes  against  the  tank. 

The  resultant  of  Ti  and  T2  is  shown  in  Fig.  31,  and  the  triangle 
aVh'  in  Fig.  31  is  similar  to  the  triangle  aOb  in  Fig.  30.  There- 
fore we  have 

r  :ds::T  :dF 
so  that 

?  =  ^  (1) 

ds       r 

We  have  already  proceeded  to  the  limit  in  considering  aOh  to 
be  a  triangle,  that  is,  in  considering  ds  to  be  a  straight  line, 
which  is  only  true  when  ds  is  infinitely  small. 

The  practical  meaning  of  equation  (1)  can  be  best  understood 

dP 
by  thinking  of    -p    as  the  really  important  quantity  under  con- 
sideration, not  by  thinking  of  the  meaning  of    F* 

*  Indeed  one  gets  into  unfamiliar  vector  theory  when  one  tries  to  consider 
the  meaning  of  F. 


CHAPTER  III. 
INTEGRATION. 

52.  Integration  as  an  arithmetical  argument. — ^Any  argument 
which  leads  to  a  proposition  concerning  total  change  when  a  rate 
of  change  is  given  may  properly  be  called  integration.  Thus  the 
following  arguments  are  the  simplest  existing  examples  of  inte- 
gration: 

(a)  A  man  earns  money  at  the  constant  rate  of  two  dollars  per 
day.  Therefore  the  man  earns  two  dollars  each  day,  and  in 
100  days  he  would  earn  200  dollars.  Or,  to  get  the  amount  of 
money  earned  in  100  days  multiply  two  dollars  per  day  by  100 
days.* 

(h)  One  man  A  earns  money  twice  as  fast  as  another  man  B. 
Therefore  in  any  interval  of  time  whatever  A  earns  twice  as 
much  as  B. 

53.  Integration  as  a  mechanical  process.    Integrating  machines. 

— The  simplest  and  most  familiar  integrating  machine  is  the 
ordinary  cyclometer.  The  wheel  of  a  bicycle  turns  at  a  speed  which 
is  proportional  to  the  velocity  of  travel  of  the  bicycle.  Therefore 
the  number  of  revolutions  of  the  wheel  is  proportional  to  the 
distance  traveled,  and  the  cyclometer  is  a  revolution  counter 
arranged  to  iesA  one  additional  unit  for  each  mile  traveled. 

The  ordinary  "electricity  meter "f  which  is  installed  in  con- 
nection with   electric   lamps   is   an   integrating  machine.    The 

*  If  one  can  get  200  dollars  by  multiplying  two  dollars  per  day  by  100  days, 
one  might  well  ask  how  to  perform  such  a  profitable  operation?  The  answer 
is:  Work  hard  for  100  days!  That  is  what  this  particular  multiplication 
means.  Nearly  every  mathematical  process  rightly  understood  relates  to 
physical  reality  in  a  manner  as  definite  and  as  exacting  as  this  particular  case 
of  multiplication. 

t  Properly  called  a  watt-hour  meter. 

74 


1-lfer- 


INTEGRATION.  75 

spindle  of  the  meter  turns  at  a  speed  which  is  proportional  to  the 
rate  of  delivery  of  energy  (in  watts)  to  a  customer.  Therefore  the 
number  of  revolutions  of  the  spindle  is  proportional  to  the  total 
amount  of  energy  delivered.  The  clock-work  of  the  watt-hour 
meter  is  a  revolution  counter  which  is  arranged  to  read  one  addi- 
tional unit  for  each  watt-hour  of  energy  delivered. 

54.  The  planimeter.* — The  planimeter  is  an  integrating  machine 
for  measuring  area,  and  it  depends  upon  the  following  kinematical 
principle : 

Consider  a  line  AB,  Fig.  32,  moving  in  any  manner  whatever 
in  the  plane  of  the  paper.  The  motion  of  the  line  at  each  instant 
may  be  thought  of  as  a  motion  of  translation  combined  with  a  motion 
of  rotation  about  an  arbitrary  point   p.f 

Let  V  be  the  component  at  right  angles  to  AB  of  the  velocity 
of  translation,  as  shown  in  Fig.  32,  and  let  to  be  the  angular 
velocity  of  AB  about  the  point  p.  It  is  desired  to  find  the  rate 
at  which  the  line  AB  sweeps  over  area,  area  swept  over  by  the 
line  being  considered  as  positive  when  the  line  sweeps  over  it  from 
right  to  left  to  an  observer  looking  from  A    towards  B. 

During  a  short  interval  of  time,  A^,  the  motion  of  translation 
carries  the  line  from   AB   to   A'B'    as  shown  in  Fig.  34,  and  the 

*  The  planimeter  which  is  here  described  is  the  Amsler  planimeter. 

The  student  wishing  to  become  familiar  with  various  kinds  of  integrating 
machines  should  consult  Les  Integraphes  Abdank-Abakanowicz,  Paris,  Gauth- 
ier-Villars. 

A  discussion  of  integrating  machines  is  also  given  in  Encyclopddie  der  Mathe- 
matischen  Wissenschaften,  Vol.  II.  This  great  work  is  now  appearing  in  a  re- 
vised form  in  a  French  translation. 

See  also  Report  on  Planimeter s,  British  Association  Annual  Report,  1894, 
pages  496-523. 

See  also  Mechanical  integrators,  Shaw,  Proceedings  of  the  Institution  of  Civil 
Engineers,  London,  1885,  pages  75-143. 

The  earhest  type  of  integrating  machine  for  use  in  harmonic  analysis  is 
that  of  Lord  Kelvin.  This  machine  is  described  in  Franklin's  Electric  Waves, 
pages  240-242,  The  Macmillan  Co.,  New  York,  1909. 

t  See  Franklin  and  MacNutt's  Mechanics  and  Heat,  Arts.  74-77,  pages 
174  and  175,  The  Macmillan  Co.,  N.  Y.,  1910. 


76 


CALCULUS. 


area  swept  over  by  the  line  is  I  X  v-At.  Dividing  this  area  by 
At  gives  the  rate  at  which  the  Une  sweeps  over  area  because  of  its 
motion  of  translation. 


/    Sf  center  of 
/   <   I 


Fig.  32. 


During  a  short  interval  of  time,    A<,    the  motion  of  rotation 
carries  the  Une  from  AB  to  A"B" ^  as  shown  in  Fig.  35,  and  the 


Fig.  35. 

\t  ■ 
area  swept  over  consists  of  two  parts  as  shown.    The  positive 


area 


\(l  \2 

in  Fig.  35  is  equal  to  2(2 +  '^)   X  w-A<,    the  negative 


area 


in  Fig.  35  is  equal  to 


\%-) 


X  oo'At,  and  the  total  area  swept 


over  is  equal  to  IDo)  •  At.    Dividing  this  area  by  At  gives  the  rate 
at  which  the  line  sweeps  over  area  because  of  its  motion  of  rotation. 


INTEGRATION.  77 

Therefore  the  total  rate  at  which  the  line  sweeps  over  area 


/dA\    . 


is  equal  to   Iv  +  IDoo,   or,  using  -77   for  w,   we  have 


Let  AB  be  a  metal  bar  carrying  a  wheel  mounted  as  shown  in 
Fig.  36  (which  is  a  side  view),  and  let  the  rim  of  this  wheel  roll  on 
the  paper  as  the  bar    AB    moves,  as  shown  in  Figs.  36  and  37. 

y^middle  of  bar  W^u„^  ao 


"\|— 4— U      paper 


j.,,roUinff 
I *  \      '^y/  wheel 


side  view  paper 

Fig.  36.  Fig.  37. 

Let  ^  be  the  total  angle  turned  by  the  wheel  in  a  given  time;  then 
^-  is  the  angular  velocity  of  the  rolling  wheel,  and  it  is  equal  to  -, 
where  r  is  the  radius  of  the  wheel,  as  shown  in  Fig.  37.  There- 
fore we  may  substitute  r  -^  for  v  in  equation  (1)  and  we  have: 

in  which    Ir  -^    is  the  rate  of  sweeping  over  area  because  of  the 

do 
translatory  motion  of  AB,  and  ID  -r.  is  the  rate  of  sweeping  over 

area  because  of  the  rotatory  motion  of  AB. 

Now  if  the  rate  of  sweeping  area  due  to  the  translatory  motion 
of  AB  is  Ir  times  the  rate  of  turning  of  the  wheel,  then  the 
total  area  swept  over  because  of  the  translatory  motion  of  AB  is 
Ir  times  the  total  angle  rp  turned  by  the  wheel. 

Similarly,  if  the  rate  of  sweeping  area  due  to  the  rotatory  motion 


78 


CALCULUS. 


of  AB  is  ID  times  the  rate  of  turning  of  AB,  then  the  total 
area  swept  over  because  of  the  turning  of  AB  is  ID  times  the 
total  angle   6  turned  by  AB. 

Therefore  the  total  area  swept  over  by    AB    is    Irrj/  +  IDd.* 
That  is, 

A  =  H-\-  IDS  (3) 

If  the  bar  AB  comes  back  to  its  initial  position  without  turning 
completely  round,  then  0  =  0  and  equation  (3)  becomes 


A  =^lri 


(4) 


That  is,  the  total  area  swept  over  by   AB   is  proportional  to  the 
angle    ^    turned  by  the  rolling  wheel,  and  the  value  of    Ir    may 

be  so  chosen  that  one  square  inch 
corresponds  to  each  revolution  of 
the  wheel.  In  this  case  square 
inches  of  area  are  indicated  by 
whole  revolutions  of  the  wheel, 
and  the  circumference  of  the  wheel 
may  be  divided  so  as  to  indicate 
fractions  of  a  square  inch. 

In  the  actual  planimeter  one  end 

of  the  bar    AB    is  constrained  to 

move  back  and  forth  along  a  circlef 

CC,     Fig.  38,  while  the  other  end 

A    travels  once    round  the  closed 

curve  CC  which  bounds  the  area  to  be 

measured.     Then  any  area  outside  of 

CC  which  is  swept  over  by  AB  at  all  is  swept  over  as  many  times 

to  the  left  as  to  the  right,  but  every  portion  of  the  shaded  area  is 

swept  over  once  more  to  the  left  than  to  the  right.     Therefore, 

*  How  much  more  easily  miderstood  is  this  argument  than  to  say  integrating 
equation  {2)  we  have  equation  {3) !  and  yet  this  brief  statement  means  exactly 
what  is  given  in  the  above  argument. 

t  Along  a  straight  line  in  some  forms  of  planimeter. 


Fig.  38. 


.m^ 


INTEGRATION. 


79 


in  view  of  the  agreement  as  to  algebraic  signs,  the  total  area 
A  (=  lr\J/)  swept  over  by  the  line  AB  is  equal  to  the  shaded  area. 

55.  Integration  by  steps. — In  the  laying  out  of  a  new  electric 
railway  the  engineer  usually  makes  a  study  of  the  probable 
schedule  speed  of  the  car  in  order  to  be  able  to  estimate  the  traffic 
income  and  to  be  able  to  choose  suitable  driving  motors.    In  the 


i€ 


«  8 


1^' 

— 

/ 

"*^ 

r^ 

/" 

^ 

/ 

/ 

■^ 

/ 

/ 
/ 

9' 

V 

/ 

1/  ^^ 

1 

24 


4  •«: 


10      20     30     40     50     60      70     80     <y3 

time  in  seconds 

Fig.  39. 

study  of  a  particular  run  of  a  car,  the  curve  A,  Fig.  39,  is  deter- 
mined from  the  known  characteristics  of  a  motor*  (tentatively 
chosen),  the  weight  of  the  car,  and  the  curvature  and  grade  of 
the  track.  The  curve  A  gives  the  speed  of  the  car  as  a  function 
of  elapsed  time  and  it  is  called  a  speed-time  curve.  In  the  particular 
case  represented  in  Fig.  39  the  car  is  being  accelerated  from  0  to 
45  seconds;  at  45  seconds  the  power  is  turned  off  and  the  car  is 
allowed  to  coast;  and  at  78  seconds  the  brakes  are  applied  and  the 
car  is  rapidly  brought  to  rest.  The  ordinate  of  curve  A  is  the 
velocity    v    of  the  car,  the  product    v-dt   is  the  distance  traveled 

during  the  infinitesimal  interval    dt,    and     I      v-dt   is  the  total 

distance  traveled  by  the  car  from  the  start  up  to  any  given  instant 
*  These  characteristics  are  furnished  by  the  manufacturing  company. 


80  CALCULUS. 

L    This  distance  is  shown  as  a  function  of  t  in  the  accompanying 

SPEED  AND  DISTANCE  TABLE  OF  AN  ELECTRIC  STREET  CAR 


(See  Fig.  39). 

Time  in 

Speed:  Miles 
Fer  Hour. 

DiBiance 

Seconds. 

.      In  Feet 

0 

0 

0 

10 

15.2 

111 

20 

22.0 

384 

30 

23.6 

719 

40 

24.2 

1,071 

45 

24.3 

1,246 

60 

21.6 

1,751 

78 

19.4 

2,292 

91 

0 

2,476 

table  and  also  it  is  shown  by  the  ordinate  of  the  curve  BB  in 
Fig.  39.  The  tabulated  values  of  the  distance  traveled  by  the 
car  are  calculated  approximately  as  follows :  From  0  to  10  seconds 

the  average  speed  of  the  car  is  approximately  ^ — -    miles  per 

hour,  and  the  distance  traveled  during  this  10-second  interval  is 
found  by  multiplying  the  average  speed  (reduced  to  feet  per 
second)  by  10  seconds  which  gives  111  feet.  From  10  seconds 
to  20  seconds  the  average  speed  of  the  car  is  approximately 

— ■ — ^ —     miles  per  hour,  and  the  distance  traveled  during 

this  10-second  interval  is  found  by  multiplying  the  average  speed 
(reduced  to  feet  per  second)  by  10  seconds,  which  gives  273  feet, 
so  that  the  total  distance  traveled  during  20  seconds  is  111  feet  + 
273  feet,  which  is  384  feet;  and  so  on. 

PROBLEMS. 

1.  The  accompanying  figure,  pi,  shows  the  stretching  force 
per  square  inch  of  section  and  the  fractional  elongation  of  a       ] 
sample  of  steel  under  test.     The  sample  is  one  inch  in  diameter       ] 
and  the  portion  which  is  stretched  is  4  inches  long  initially.     Find 


INTEGRATION. 


81 


the  work  in  foot-pounds  expended  on  the  specimen  up  to  the 
breaking  point   h. 
Ans.    53.1  foot-pounds. 

Note. — Multiplying  abscissas  in  Fig.  pi  by  4  inches  gives  actual  elongations 
in  inches,  and  multiplying  ordinates  by  square  inches  of  section  of  sample 
gives  actual  stretching  forces  in  pounds.  Let  elongation  in  inches  be  e  and 
let  stretching  force  be  F.    Then  work  done  is  the  integral  of  F-de. 

2.  Find  the  work  (in  ergs  and  in  foot-pounds)  required  to 
magnetize  a  wrought  iron  bar  3  inches  by  3  inches  by  20  inches 


<s 

m  <tm 

■■ 

■■ 

■■ 

~ 

■" 

" 

■" 

"" 

■" 

"■ 

■■ 

"" 

■" 

"" 

"■ 

- 

- 

' 

1^ 

— 

■* 

^' 

1 

■^ 

.^ 

/- 

^ 

J 

X 

^ 

\t 

,  ' 

s 

s 

^ 

y 

1 

s 

/ 

^ 

\ 

1 

/ 

. 

« 

/ 

\\ 

S; 

f>    J 

f 

P 

y 

v 

/\ 

- 

Kr 

^ 

? 

' 

^45 

g 

/ 

§ 

/ 

Vj 

<. 

/ 

B 

^•'" 

/ 

s 

/ 

g 

/ 

s 

/ 

«7r 

/ 

s 

/ 

i 

/ 

% 

/^ 

/ 

/ 

_ 

5,  JO  15  .20  25  30 

ebieUsi,  bandredtba  tot  curve  A,  tea-tbouaandUi9  for  curv»  B- 

Fig.  pi. 

from  a  neutral  condition  to    B  =  16,000    lines  per  square  centi- 
meter. 

Ans.    326  X  10^  ergs  or  2.406  foot-pounds. 

Note, — ^The  work  done  in  magnetizing  iron  is: 


'^  =  If^ 


dB 


82  CALCULUS. 

where  W  is  expressed  in  ergs  and   V  is  the  volume  of  the  iron  in  cubic  cenf  i- 
meters.    There  are  30.5  X  453.6  X  980  ergs  in  one  foot-pound. 

TABLE  OF  B  AND  H  FOR  WROUGHT  IRON. 

H  B 

10  12,400 

20  14,330 

30  15,100 

40  15,550 

50  15,950 

60  16,280 

70  16,500 

3.  The  force  F  which  acts  on  a  body  is  given  as  a  function  of 
the  time  in  the  accompanying  table.  Force  being  expressed  in 
poundals  and  time  in  seconds.  The  mass  of  the  body  is  1,000 
pounds.  Find  the  velocity  produced  from  ^  =  0  to  i  =  40 
seconds. 

Ans.    27.1  feet  per  second. 

dv        F  F 

Note. — The  acceleration  -j-.  =  —  or  dv  =  —'dl  or  the  velocity  v  is  equal 
dt      m  m  J  -i 

to  —   times  the  integral  of  F-dt;  velocity  being  expressed  in  feet  per  second. 

TABLE  OF  F  AND  L 

t  in  Second!.  F  in  Poundali. 

0  500 

10  625 

20  700 

30  745 

40  785 

56.  Algebraic  integration. — Integration,  in  the  sense  in  which 

this  word  is  used  throughout  this  text,  is  the  recognition  of  a 

given  differential  expression  as  the  differential  of  a  known  function, 

and  the  rules  for  differentiation  which  are  given  in  Chapter  II 

are  rules  for  integration  also.     Thus  the  differential  of  log  x  +  C 

dx  dx 

is  — ,  and  therefore,  by  definition,  the  integral  of  —  is  loga;+C, 

X  X 


INTEGRATION.  83 

where  C  is  any  constant.  The  rules  given  in  Chapter  II  cover 
differentiation  completely,  or  nearly  so,  but  they  do  not  cover 
integration  completely.  Indeed  complete  rules  cannot  be  given 
for  integration.* 

In  the  recognition  of  a  given  differential  expression  as  the 
differential  of  a  known  function,  two  things  are  helpful,  namely, 
(a)  A  table  of  standard  forms  or  functions  with  their  differentials 
(such  a  table  is  usually  called  a  table  of  integrals),  and  (6)  some 
degree  of  familiarity  with  the  transformations  which  may  be 
used  to  reduce  a  given  differential  expression  to  a  standard  form 
or  to  a  combination  of  standard  forms. 

A  table  of  integrals  is  given  in  Appendix  B. 

57.  Transformation  of  differential  expressions. — When  a  differ- 
ential equation  has  been  found  by  differentiating  a  given  function, 
the  integral  of  the  differential  equation  is  of  course  known.  Many 
of  the  forms  in  the  table  of  integrals  have  been  established  in 
this  way  as  problems  in  differentiation  in  Chapter  II.  When  a 
given  differential  equation  has  not  been  found  by  differentiating 
a  known  function  it  is  in  many  casesf  possible  to  invent  a  scheme 
for  transforming  the  given  differential  expression  so  as  to  reduce 
it  to  one  or  more  of  the  standard  forms  in  the  table  of  integrals. 
The  most  important^  of  these  transformations  are  embodied  in 
the  following  rules,  and  a  few  of  the  lesser  important  transforma- 
tions are  illustrated  by  problems. 

As  a  very  simple  example  consider  the  differential  equation 

*  Integration  by  series  is  a  universal  and  systematic  method  of  integration. 

dv 
Any  given  algebraic  expression  for    -^     can  be  expanded  by  Maclaurin  s 

theorem,  each  term  of  the  series  so  obtained  can  be  integrated  according  to 
form  1  in  the  table  of  integrals,  and  the  resulting  series,  when  it  is  convergent 

is  the  integral  of  the  given  expression  for  -p . 

t  Not  in  every  case,  by  any  means;  many  integrals  cannot  be  expressed  in 
terms  of  the  elementary  functions  which  are  used  in  the  table  of  integrals. 

X  Transformations  depending  upon  the  use  of  imaginaries  are  also  very 
important.     See  Art*  93  in  Chap.  VI. 


84  CALCULUS. 

which  was  found  in  the  discussion  of  the  arc  of  a  parabola  in 
Art.  20,  namely: 

ds  =  VT+lfc2^2 .  dx  (1) 

This  expression  is  easily  reduced  to: 

ds=2ky]^,-hx''dx  (2) 

and  according  to  rule  I,  below,  we  have: 

h^li'  +  *=  •  dx  =  2kf^±  +x^.dx  (3) 

Therefore,  writing    a^    for    — ,    we  get  an  expression  which  is 

identical  to  form  57  in  the  table  of  integrals. 

Rule  I. — Any  constant  factor  may  be  removed  from  under  the 
integral  sign.    Thus: 

J  a  '  du  =  ajdu 

where  a  is  a  constant  and  u  is  any  differential  expression  what- 
ever. 

Rule  n. — The  integral  of  the  sum  of  two  differential  expressions 
is  equal  to  the  sum  of  the  integrals  of  the  respective  expressions. 
Thus: 

J(du  +  dv)  =  f  du  -\-  jdv 

where  du  and  dv  are  any  differential  expressions  whatever. 

Rule  in. — When  a  differential  expression  is  of  the  form  w"  •  du 
its  integral  is,  of  course,  given  by  form  1  or  by  form  2  of  the  table 
of  integrals,  whatever  the  function  u  may  be. 

For  example  consider  the  differential  equation: 

dy  —  sin^  x  cos  x  *  dx  (4) 

Let    mix  =  u    then    sin^  x  =  u^    and    co^x  •  dx  =  duj    and 
equation  (4)  becomes: 

dy  =  u^  •  du 


."ii^ii- 


INTEGRATION.  85 

whence 

y^iu'  +  C 

or,  substituting  sin  a;  for  u  we  have. 

y  =  i  sin^  x  -j-  C  (5) 

For  problems  see  group  7  in  the  appendix. 

Rule  IV.    Integration  by  parts. — This  rule  is  expressed  thus: 

fu'dv  =  uv  —fv  •  du  (6) 

This  formula  is  sometimes  useful  when  a  given  differential  ex- 
pression can  be  resolved  into  two  factors,  namely,  (a)  any  function 
u,   and  (6)  a  familiar  differential  expression  dv. 

This  rule  is  based  on  Art.  32  where  it  is  shown  that: 

d(uv)  =  u  •  dv  -\-  V  '  du 
which  is  to  say: 

uv  =fu  •  dv  +fv  '  du 

so  that,  by  transposing  we  get  equation  (6). 
For  example  consider  the  differential  equation: 

dy  =  X  log  X  '  dx  (7) 

In  this  case  x  -  dx(=  dv)  is  a  familiar  differential  expression, 
that  is,  it  is  the  differential  of  ix^{=  v).  Therefore,  using  equa- 
tion (6),  we  have: 

J       u    •    dv       =       u    '  v      —f    V  '    du  (6) 

f{\ogx)(x'dx)  =  {\ogx)(ix')-fiix')(j)  (8) 

But  (ia;2)  f  —  j  is  Jx  •  dx,  and  its  integral  is  ix^  +  C.  Therefore 
equation  (8)  becomes: 

fx  log  X  '  dx  =  ix^  log  X  -  ix^  -  C  (9) 

For  problems  see  group  8  in  the  appendix. 


86  CALCULUS. 

Special  methods.  Some  differential  expressions  require  more 
or  less  elaborate  transformations  or  substitutions  to  get  them  into 
forms  whose  integrals  can  be  recognized,  or  easily  obtained  by  the 
simple  transformations  of  rules  III  and  IV  above.  Among  these 
special  methods  are  the  following: 

(a)  Integration  of  rational  fractions  by  resolution  into  partial 
fractions.  For  'problems  illustrating  this  method  see  Group  9  in  the 
Appendix.  

(6)  Integration  of  expressions  involving  Va^  —x  ^,  V  a;^  +  a* 
or  V  x^  —  a^.  Such  expressions  can  usually  be  reduced  to  recog- 
nizable ^ormsJ)y  making  the  following  substitutions. 

In  Va^  —  x^  substitute  a  sin  d  for  x. 

In  V^^^J^  substitute  a  tan  d  for  x. 

In  Vx^  -f  a^  substitute  a  sec  $  for  x. 

Problems  illustrating  these  trigonometric  substitutions  are  given  in 
Group  10  of  the  Appendix. 

(c)  Miscellaneous  Substitutions.  A  few  problems  involving  mis- 
cellaneous substitutions  are  given  in  Group  11  in  the  Appendix. 

A  good  discussion  of  integration  transformations  is  given  on 
pages  32-70  of  W.  E.  Byerly's  Integral  Calculus^  Ginn  &  Co., 
Boston,  1881 J  second  edition^  1^ 


CHAPTER  IV. 
PARTIAL  DIFFERENTIATION  AND  INTEGRATION. 

58.  Differentiation  of  a  function  of  two  variables. — Consider  a 
rectangle  of  length   x  and  breadth   y   as  shown  in  Fig.  40.    The 

area  of  the  rectangle  is: 

A  =  xy  (1) 

that  is,  the  area  is  a  function  of  the  two  variables  x  and  y. 
Writing  A  -\-  AA   for  A, 

X  +  Ax    for  X,    and 

y  -\-  Ay    for  y,    we  have: 

A  +  AA  =  (x  +  Ax){y  +  Ay) 
or 

A-\-AA='xy-\-X'Ay-\-y'Ax-{-Ax'Ay  (2) 

whence,  subtracting  equation  (1)  from  equation  (2)  member  by- 
member,  we  have 

AA  ==^  X  '  Ay  ■\-  y  '  Ax  ■\-  Ax  ^  Ay  (3) 

But  when    Ax    and    Ay    are  infinitely  small  we  may  drop  the 
second  order  infinitesimal  Ax  •  Ay,  and 
we  get: 

dA  =  X  •  dy  ■{•  y  '  dx  (4) 

Now  X  '  dy  is  the  infinitesimal  incre- 
ment of  A  when  y  alone  increases, 
and  y  '  dx  \&  the  infinitesimal  incre- 
ment of  A  when  x  alone  increases. 
Therefore,  according  to  equation  (4),  the  infinitesimal  increment 
of  A  when  x  and  y  both  increase  is  equal  to  the  sum  of  the  two 
infinitesimal  increments: — (a)  that  which  is  due  to  the  increase  of  x 
alone  and  (6)  that  which  is  due  to  the  increase  of  y  alone. 

87 


88  CALCULUS. 

This  proposition  is  a  special  case  of  a  general  theorem  which  is 
expressed  by  the  equation: 

dz 
where   —  is  the  derivative  of  z  with  respect  to  x  on  the  assump- 

dz 
tion  that  y  is  constanty  and  —   is  the  derivative  of  z  with  respect 

dz 
Uyyonthe  assumption  that  x  is  constant;  that  is  -^  *  dx  is  the 

ox 

infinitesimal  increment  of    z    due  to  an  increment  of    x    alone, 

IT  '  dy   is  the  infinitesimal  increment  of   z  due  to  an  increment 
dy 

of    y    alone,  and    dz  I  =-r~ '  dx  +  —  -  dy  )     is  the  infinitesimal 

increment  of  z  when  x  and  y  both  increase. 

It  is  customary  to  use  the  symbol  d  instead  of  d  in  partial 
differentiation,  but  to  do  so  is  apt  to  be  misleading  because  it 
gives  the  impression  that  there  is  an  unexplained  and  mysterious 
difference  between  ordinary  and  partial  differentiation.  It  is 
indeed  allowable  to  use  d  instead  of  d  because  one  always  knows 
when  one  is  dealing  with  a  function  of  more  than  one  independent 
variable. 

59.  Successive  partial  differentiation. — Let  y  he  a.  function 
of  a  single  independent  variable  x.  Then  the  rate  of  change  of  y 
with  respect  to  x  is  called  the  derivative  of  y  with  respect  to  x. 
But  this  derivative  is  itself  a  function  of  x  and  its  rate  of  change 
with  respect  to  x  is  called  the  second  derivative  of  y  with  respect 
to  x;  and  so  on  as  explained  in  Art.  46. 

Let   2    be  a  function  of  two  independent  variables   x    and    y. 

Then  the  two  first  derivatives  r-  and  t-  are  themselves  functions 

ax  dy 

of  x  and  y,  and  each  of  these  first  derivatives  has  its  derivative 

with  respect  to    x    and  its  derivative  with  respect  to    y.    For 


PARTIAL  DIFFERENTIATION  AND  INTEGRATION.        89 

example,  if  2  =  T^y^y  then   —  =  3xy(=a),  and  —  =3a;y(=/3); 

and  each  of  these  first  derivatives  is  a  function  of  x  and  y.    The 
•\  »\ 

derivative  t~  (=  ")    ^^s  two  derivatives,  namely,    t-  (  =  6x2/^) 
ox  ox 

and  ^(=  OxV);  and  the  derivative   —  (=  /S)  has  two  deriva- 
tives, namely,  ^  (=  9xV)   and   7-  (=  6x^1/), 

dz  d^Z 

The  derivative  of  —  with  respect  to  x  is  indicated  as     r^. 

62  d^Z 

The  derivative  of  —  with  respect  to  y  is  indicated  as —. 

uX  oy  '  ox 

dz  d^z 

The  derivative  of  —  with  respect  to  x  is  indicated  as  -r r-. 

dy  ^  dx  '  dy 

The  derivative  of  —  with  respect  to  y  is  indicated  as  r-r. 

In  the  above  example,  namely,  when   z  =  oi^y^j  the  two  second 

d^z  d^z 

derivatives,   -r-^-    and  r-r-,    are  each  equal  to    9xV.    That  is 
dydx  dxdy  ^  ^ 

these  two  second  derivatives  are  identical.    Indeed  the  two  deriva- 
tives T-^    and  ^-T-    are  always  identical  when  z  is  a  function  of 

X    and    y.    This  proposition  is  very  important  in  mathematical 
physics. 

60.  Partial  differential  equation.* — Let    2   be  a  function  of   x 
and  y  concerning  which  it  is  known  that 

1=  ax^  (1) 

and  let  it  be  required  to  find  an  expression  for    z.    Now    y    is 

dz 

assumed  to  be  a  constant  when  the  derivative  r-  is  found  from  a 

OX 

*  A  very  interesting  example  of  a  partial  differential  equation  is  discussed 
in  Art.  90. 


90 


CALCULUS. 


given  function,  and  therefore  y  is  to  be  looked  upon  as  a  constant 
in  equation  (1)  when  one  tries  to  think  of  the  function  from  which 
equation  (1)  is  derived,  that  is,  ay  in  equation  (1)  is  to  be  con- 
sidered as  a  constant,  so  that  we  may  write 


ay  =  h 


(2) 


and  consider  what  function  of  x  has  a  derivative  equal  to  hx^. 
Evidently  the  desired  function  of  x  is  [^bx^  +  C].  But  in  this 
argument  y  is  assumed  to  be  constant,  so  that  the  constant  of 
integration  C  can  be  any  function  whatever  of  y.  Therefore, 
writing  f(y)  for  C  and  using  ay  for  h,  we  have  Hayx^ -{- f{y)] 
as  the  most  general  algebraic  expression  which  gives  ax^y  when 
differentiated  with  respect  to  x,  and  this  is  therefore  the  desired 
expression  for  z.    That  is: 

z  =  iai^y+f{y)  (3) 

where  f(y)  represents  any  function  of  y  whatever.  Of  course 
this  function  f{y)  may  include  a  term  which  does  not  contain  y, 
as  in  ay^  +  hy  +  c,  or  indeed  it  may  be  a  simple  constant  which 
does  not  contain  y  at  all. 

Equation  (1)  expresses  the  law  of  growth  of  z  with  respect 
to  X,  and  it  is  called  a  partial  differential  equation  because  z  is 
a  function  of  more  than  one  independent  variable. 


Example  of  an  ordinary  differ- 
ential equation. 


If 


then 


dz 
dx 


ax^ 


(4) 


(5) 


z==iax^-{-C 

where    C    is  a  constant  which 

may  have  any  value  whatever. 

If  we  place  x  =  0  inequation 


Example   of  a   partial   differ- 
ential eqvxition. 


If 


then 


52 

dx 


=  ax"^ 


(6) 


z^\a:>?-\-f{y)         (7) 

where  f{y)  is  any  function 
whatever  of  y  which  does  not 
depend  on  x. 


PARTIAL  DIFFERENTIATION  AND  INTEGRATION. 


91 


(5)  we  have  z  =  C.  Therefore 
the  constant  C  is  determined 
if  we  know  the  value  of  z  when 
a;  =  0. 

The  constant   C  is  called  the 
\  constant  of  integration. 


If  we  place  x  =  0  in  equa- 
tion (7)  we  have  z  =  f{y). 
Therefore  the  function  f{y)  is 
determined  if  we  know  2  as  a 
function  of  y  when  a;  =  0. 

The  function  f{y)  is  called 
the  function  of  integration. 


The  above  discussion  refers  to  very  simple  partial  differential 
equations.  As  another  example  let  us  set  up  the  partial  differ- 
ential equation  which  characterizes  a  given  function.  Let  z  be 
any  function  whatever  of  s  where  s  stands  for  x  +  ay.  Then 
ds 
dx 


=  V 


,   ds 
and  -y-  =  a. 
dy 


Also,  according  to  Art.  34,  we  have: 


dz 
dx 


dz 
ds 


-r  =  -r  •  -7-         and         -r  =  -i- 


dx 


Therefore,  using  the  values, 
dz 


ds 
dx 


=  1   and 


d2 
dx 


and 


dz  _  dz 
dy  ds 
ds 
dy 
dz  dz 
dy 


dy 


=  a. 


we  have 


ds 


and  consequently: 


dz 

dy 


=  a 


dz 

dx 


(8) 


This  is,  of  course,  a  partial  differential  equation,  and  it  is  satisfied 
by  any  function  whatever  of  {x  -\-  ay).  For  example  it  is  satis- 
fied by  z  =  X  -{-  ay,  by  z  =  {x  -\-  ayy,  by  sin  {x  +  ^y),  by 
log  {x  +  ay),   by   e^^^',   etc. 

6L  Differentiation  of  an  implicit  function. — Any  equation 
between  x  and  y  defines  2/  as  a  function  of  x^.  If  the  equa- 
tion is  solved  for  y,  we  have  y  as  an  explicit  function  of  x; 
otherwise  the  equation  defines    y    as  an  implicit  function  of    x. 

*  In  this  discussion  the  symbol  d  is  used  instead  of  d  for  partial  differentia- 
tion. 

t  Or  X  as  a  function  of  y. 


92  CALCULUS. 

Thus 

6X2  ^  IQ^yS  +  2/4  +  5  ==  0  (1) 

defines  y  as  an  implicit  function  of  x.  It  is  convenient  to  repre- 
sent the  entire  left-hand  member  of  equation  (1)  by  the  letter  /. 
Imagine  for  a  moment  that  /  need  not  be  equal  to  zero  as  required 

by  equation  (1),  then   ^  •  dx  would  be  the  increment  of  /  due 

to  an  increment  of  x,  and   ^  •  dy  would  be  the  increment  of  / 

due  to  an  increment  of  y.  But  according  to  equation  (1)  /  is 
always  equal  to  zero.  Therefore  the  increments  dx  and  dy 
must  be  so  related  to  each  other  as  to  make  the  sum  of  the  two 
increments  of  /  equal  to  zero.    That  is  we  must  have: — 

)  (2) 


(3) 


when    y    is  an  implicit 

function  of  a;  as  defined  by  the  equation  /  =  0,  where  /  is  any 
algebraic  expression  involving  x  and  y. 

In  the  above  expression  the  two  differentials  dx  and  d-^  are 
called  increments  for  the  sake  of  brevity  and  clearness.  A  differ- 
ential is  of  course  not  an  increment  because  it  is  infinitely  small. 

PROBLEMS. 

dv 
Using   the   method    of   Art.    61,    find     -^     in  the   following 

expressions. 

1    ?!_L.^'_i  =n  dy^_^ 

*•  a^'^b^  ^'  dx  a'y 


' 

df 
dx 

dx  + 

dy 

'dy 

or,  solving  for 

dy 
dx' 

we  have 

dy 

dx 

d^  " 

^1 
dy 

This  formula  gives  the  value  of 

dy 
dx 

dy 
dx 
dy 

x^  -  ay 
'  ax  -  t' 

6a: +  52^ 

dx 
dy 

y\\hx  +  22/) 
x^  +  y^x^  -  y^ 

dx 

dy  _ 
dx 

x{y  +  Vx2  -  2/2) 
y  -  X 
'  y  +  x' 

PARTIAL  DIFFERENTIATION  AND  INTEGRATION.        93 

2.  a:^  +  2/'  -  ^axy  -3  =  0, 

3.  Qx^  +  lOxy^  +  2/'  +  5  =  0, 

4.  log  (x^  -  2/2)  -  2  sin-i  -  =  0, 

X 

5.  2  tan-i  -  -  log  {x^  +  y^)  =  0, 

y 

62.  Slope  of  a  hill. — Imagine  a  hill  built  upon  the  plane  which 
contains  the  x  and  y  axes  of  reference,  and  let  z  be  the  height 
of  the  hill  above  the  point  on  the  plane  whose  coordinates  are 
X  and  y.  Then  an  equation  expressing  z  as  a  function  of  x 
and  2/  is  the  equation  of  the  surface  of  the  hill.     Thus 

z  =  Vr2  -  a:2  -  2/2  (1) 

is  the  equation  of  a  hemispherical  hill  of  radius  r. 

Consider  the  derivatives  -r-  and  r- ;  these  derivatives  are  both 
dx  dy  

we  have 


functions  of  x   and   y.    Thus  if  2  =  Vr^  —  x^  - 
dz                      X 

-f, 

dx           Vr2  -  x^  -  y^ 
and 

dz                   y 

dy           Vr2-a:2-2/2* 

Let  a:',  2/'  and  2'  be  the  coordinates  of  a  particular  point  on  the 
surface  of  the  hill  as  shown  in  Fig.  41.     If  the  values  x'   and   2/' 

are  substituted  for    x   and    y   in  the  general  expressions  for    — 

ox 

dz 
and  —  we  get  the  values  of  these  derivatives  at  the  point  p,  and  we 

will  represent  these  values  by   (  t"  )     and    (  t-  )  •    Then   (  —  1 


94 


CALCULUS. 


is  the  slope  of  the  tangent  Hne  qq  and  it  is  equal  to  tan  a  where 
a  is  the  angle  shown  in  Figs.  41a  and  416;  and   (  7-  I     is  the  slope 


\2-axi8 


>^     x^axht 


y-axis 


Fig.  41a. 


A 

~S^^ 

^ 

^S 

y 

^r 

X 

/  X-axis 

y 

'«\  / 

X' 

/f*' 

// 

>' 

/v-axi» 

Fig. 

416. 

of  the  tangent  line   rr   and  it  is  equal  to   tan  /3   where   /3   is  the 
angle  shown  in  Figs.  41a  and  416. 

In  Fig.  41a  —  and  —  are  both  negative;  that  is,  z  decreases 

as    X    increases,  and    z    decreases  as    y    increases.     In  Fig.  416 

■r-   and  -r-   are  both  positive. 
dx  dy 

63.  Equation  of  the  tangent  plane. — The  derivation  of  the 
equation  of  the  tangent  plane  which  touches  a  surface  at  a  point 
is  somewhat  similar  to  the  derivation  of  the  equation  of  a  tangent 
line  which  touches  a  curve  at  a  point.  Therefore  it  is  worth  while 
to  give  the  derivation  of  the  equation  of  a  tangent  line  before 
considering  the  equation  of  a  tangent  plane. 

Let  a;'  and  z'  be  the  coordinates  of  the  point  p  where  the 
tangent  line  touches  the  plane  curve    cc,    as  shown  in  Fig.  42a. 


PARTIAL  DIFFERENTIATION  AND  INTEGRATION. 


95 


Starting  at  the  point  p  in  Fig.  42a,  which  is  at  a  height  z'  above 
the  base  Hne,  travel  along  the  tangent  line    tt    covering  the  hori- 


Fig.  42&. 

zontal  distance    {x  —  x')    and  the  rise  will  be    {x  —  x')  tan  a   or 

dz 
(x  —  x')  -J-.    We  thus  reach  any  given  point    B    in  the  tangent 

line  of  which  the  coordinates  are  x  and  z,   and  the  value  of  z  is 
s  =  2'  +  (^  -  a;')  S  (1) 


which  is  the  desired  equation  of  the  tangent  line   tt. 

Let  x',  y'  and  z'  be  the  coordinates  of  the  point  p  where  the  tan- 
gent plane  touches  the  given  surface 
(see  Fig.  426) .  Starting  at  the  point  p, 
which  is  at  a  distance  z'  above  the 
base  plane,  travel  along  the  tangent 
line  rr  (which  lies  in  the  tangent 
plane)  covering  the  horizontal  dis-  t^ 
tance  (y  —  y')  towards  the  reader 
in  Fig.  426,  and  the  rise  will  be 


(2/-2/')tan/3=(2/-2/')(|)^ 


96 


CALCULUS. 


Then  continue  along  the  Hne  q'q'  parallel  to  the  tangent  line  qq 
(the  line  q'q'  lies  in  the  tangent  plane)  covering  the  horizontal 
distance   (x  —  x')   to  the  right  in  Fig.  426,  and  the  rise  will  be 

{x  -  x')  tan  a  =  (x  -  ^')\r£) 

We  thus  reach  any  given  point  B  in  the  tangent  plane  of  which 
the  coordinates  are  x,  y  and  z,  and  the  value  of  z  is: 


2  =  0'  + 


which  is  the  desired  equation  of  the  tangent  plane.    Of  course  the 
derivatives  in  this  equation  refer  to  the  equation  of  the  hill. 

64.  Component  slopes  and  resultant  slope. — Let    X    be  the 

slope  of  a  hill  at  the  point  p,  Fig.  41,  in  a  direction  parallel  to  the 

the  rise*  per  unit-horizontal-distance- 
parallel-to-the-x-axis,  and  it  is  equal  to 

Also  let    Y    be  the  slope  of 

the  hill  at   p   in  a  direction  parallel  to 

theiz-axis.    Then  Y  =  (^)  .     Let  X 


x-axis. 


That 


IS, 


X     is 


x-axia 


Wp 


and  Y  be  represented  to  a  chosen 
scale  by  the  lines  X  and  Y  in  Fig. 
43.  Then  the  diagonal  R  in  Fig,  43 
represents  what  may  be  called  the  result- 
ant or  actual  slope  of  the  hill  at  p. 
That  is,  the  line  R  points  directly  up 
hill,  and  the  length  of  R  represents  (to 
the  chosen  scale)  the  rise  of  the  hill  per 
unit-of-horizontal-distance-in-the-direction-of-/2.  This  is  a  funda- 
mental and  important  theorem  in  the  mathematical  theory  of 
electricity  and  magnetism,  and  the  proof  of  the  theorem  is  as 
follows: 

*  If  the  slope  is  negative,  X  is  the  drop  per  unit  horizontal  distance,  etc. 


y-axis 

Fig  43. 

The  X  and  y  axes  are  shown 
as  they  would  appear  if  seen 
from  above  in  rigs.41a  and  416. 


PARTIAL  DIFFERENTIATION  AND  INTEGRATION. 


97 


Let  the  line  AB  in  Fig.  44  be  the  line  of  intersection  of  a 
horizontal  plane  through  p  with  the  plane  which  is  tangent  to 
the  hill  at   p.     Then  the  line   AB   is  evidently  at  right  angles  to 


y-axia 


Fig.  44. 

R.  Let  the  line  ah  he  the  line  in  the  tangent  plane  at  every  point  of 
which  the  tangent  plane  is  unit  distance  above  the  horizontal  plane 
through  p.    Then,  by  definition: 


and 


R  = 


X  = 


Y  = 


pr 
1 

ps 
J_ 
pq 


(1) 
(2) 
(3) 


where   pr,   ps   and   pq   are  the  distances  shown  in  Fig.  44.    But 
1  ,  —         vr  1 


ps 


pr    _ 
cos  6       R-cosO 


,  and  pq  — 


sin  0     R  '  sin  0' 


Therefore 


and 


X  =  RcosB 
Y  =  Rsind 

Z2  +  F2  =  R^ 


(4) 
(5) 

(6) 


98 


CALCULUS. 


These  equations  establish  the  proposition  that  R  is  the  diagonal 
of  a  rectangle  of  which  X  and  Y  are  the  sides  as  shown  in  Fig.  44. 

65.  Example  of  gradient  in  two  dimensions. — The  observed 
pull  in  Fig.  10  is  a  function  of  one  independent  variable,  the  current 
as  indicated  by  the  ammeter,  and  in  such  a  case  it  is  helpful  to 
plot  a  curve  of  which  the  abscissas  represent  observed  currents 
and  of  which  the  corresponding  ordinates  represent  observed 
values  of  pull.     This  curve  is  shown  in  Fig.  11. 

Frequently  one  is  concerned  with  a  quantity  which  depends 
upon  two  independent  variables.     For  example  one  might  be 

concerned  with  the  distri- 
bution of  temperature  over 
a  flat  metal  plate.  In  such 
a  case  it  is  helpful  to  think 
of  the  temperature  at  each 
point  p  as  represented  by 
the  height  at  p  of  a  hill. 
Then  the  component  gradi- 
ents of  temperature  are  rep- 
resented everywhere  by  the 
component  slopes  of  the 
hill,  and  the  resultant  gra- 
dient of  temperature  is  rep- 
resented everywhere  by  the 
resultant  slope  of  the  hill,  as  indicated  by  the  sides  and  diagonal 
of  the  rectangle  in  Fig.  45. 

66.  Example  of  gradient  in  three  dimensions. — One  might  be 
concerned  with  the  distribution  of  temperature  throughout  a  solid 
body.  Of  course,  the  temperature  would  have  a  definite  valiie  at  each 
point  in  a  body,  or,  as  expressed  in  mathematical  language,  the 
temperature    T  at  each  point  would  be  a  definite  function  of  the 

coordinates  x,  y  and  z  of  the  point,  and  the  derivatives  ^— ,   -r- 


x-axi8 

metal 
plate 

'< 

^    i 

x" 

Y 

U^axis 

H 

Fig  45. 


and  -^r-   would  also  be  functions  of   x,   y   and   z. 
oz 


dx^    dy 
That  is,  the 


PARTIAL  DIFFERENTIATION  AND  INTEGRATION.        99 

three  derivatives  would  have  definite  values  at  each  point  in  the 

body,  and  the  three  derivatives,  --,   -^-  and  -r-,   are  the  com- 

ponent  gradients  of  the  temperature  at  the  point  in  directions 
parallel  to  the  respective  axes  of  reference.  Representing  the 
resultant  temperature  gradient  at  a  point  by  R  and  the  three 
components  of  R  by  X,    Y  and  Z,  we  have 

R^  =  X'+Y^  +  Z2  (1) 

2  =  S  (4) 

These  equations  constitute  an  extension  to  three  dimensions  of 
the  theorem  of  Art.  64. 

In  Fig.  45  the  direction  at  right  angles  to  the  surface  of  the 
plate  is  available  for  purposes  of  geometrical  representation,  and 
one  can  think  of  an  actual  hill  being  built  upon  the  plate  so  that 
the  temperature  of  the  plate  at  each  point  is  represented  by  the 
height  of  the  hill  at  that  point,  and  of  course  this  "  hill"  becomes 
a  "valley"  dipping  below  the  plate  where  the  temperature  of  the 
plate  is  below  zero. 

The  temperature  at  each  point  of  a  solid  body,  however,  cannot 
be  represented  geometrically  as  a  height  because  space  is  filled 
in  every  direction  by  the  body  itself.  That  is,  every  direction 
in  space  is  used  for  representing  the  independent  variables  x,  y 
and  z,  and  no  direction  is  left  for  the  representation  of  T.  It  is 
convenient,  however,  even  in  this  case,  to  speak  of  the  "temperature 
hill,^'  the  "height"  of  the  hill  at  each  point  of  the  body  being  the 
temperature  itself,  high  or  low  as  the  case  may  be. 

67.  Partial  integrations.* — In  many  cases  the  derivative  of  a 

*  What  is  here  called  partial  integration  is  usually  called  multiple  integra- 
tion, double,  triple,  quadruple,  etc.,  as  the  case  may  be. 


100 


CALCULUS. 


function  of  one  independent  variable  can  be  set  up  or  established 
by  the  use  of  elementary  principles  of  physics  and  arithmetic, 
and  the  function  itself  can  be  found  by  one  integration,  as  exempli- 
fied in  Arts.  22  and  23.  l 

But  when  the  function  to  be  established  is  a  function,  say,  of 
two  independent  variables  {x  and  y  for  example),  then  it  is 
usually  necessary  to  perform  an  integration  with  respect  to  x 
in  setting  up  the  derivative  of  the  desired  function  with  respect 
to  2/,  or  to  perform  an  integration  with  respect  to  y  in  setting 
up  the  derivative  of  the  desired  function  with  respect  to  x.  Such 
integrations  may  be  called  partial  integrations  because  they  are 
related  to  partial  differentiation  in  the  same  way  that  ordinary 
integration  (with  respect  to  one  independent  variable)  is  related 
to  ordinary  dijfferentiation  (with  respect  to  one  independent 
variable). 

68.  Volume  of  a  hill.    Example  of  partial  integrations. — For 

the  sake  of  simplicity  let  us  consider  a  particular  case,  namely,  a 
hemispherical  hill  with  its  center  at  the  origin  of  coordinates. 
Then  the  equation  of  the  surface  of  the  hill  is 


=  Vr2  -  x2  - 


r 


(1) 


Thus  Fig.  41a  represents  one  quarter  of  a  hemispherical  hill, 
and  it  is  required  to  find  its  volume. 


Iz-axU 


I 


^ff-axis 


Fig.  46a. 


Fig.  466. 


i 


PARTIAL  DIFFERENTIATION  AND  INTEGRATION.       101 

Let  V  be  the  volume  of  the  portion  ahcABC  of  the  hill,  as 
shown  in  Fig.  46a.  Then  y  is  a  function  of  y  (see  figure),  and 
the  increment  of  v  due  to  an  infinitesimal  increment  of  y  is  the 
volume  of  the  thin  slab  in  Fig.  466.     Therefore  we  have: 

dv  =  A  '  dy  (2) 

where   A    is  the  area  of  the  face  of  the  slab  as  shown  in  Fig.  466. 
Therefore  we  must  find  an  expression  for    A    before  we  have  a 

known  expression  for   -7-. 

Now  ape  in  Fig.  46a  is  a  plane  curve,  and  equation  (1)  is  the 
equation  of  this  curve  (ordinate  z,  abscissa  x)  if  y  is  constant. 
Let  a  be  the  shaded  area  in  Fig.  46a;  then  the  strip  with  double 
shading  is  da  and  it  is  equal  to  z  •  dx.  Therefore  using  the 
value  of  z  from  equation  (1)  we  have: 

da  =  Vr2  -  ^2  -  2/2  .  dx  ■     (3) 

and  the  total  area    A    of  the  face    abcp    is  the  integral  of  this 
expression  from  a;  =  0  to  x  =  ha  =  Vr^  —  y^*    Therefore: 

Vr2  -  x^-  ?/2  .  dx  (4) 

0 

In  performing  this  integration  y  is  a.  constant.  This  integral 
gives  an  expression  for  A,  and  the  value  of  A  so  found  can  be 
substituted  in  equation  (2).  Then  equation  (2)  may  be  integrated 
between  the  limits  y=  0  to  y  =  r  to  give  the  desired  expression 
for  the  volume   V  of  the  hill.    That  is 


v=r 

*Jy=0 


A  '  dy  (5) 


The  above  discussion  applies  to  a  particular  case,  namely,  to  a 
quarter  of  a  hemispherical  hill;  but  the  volume  of  any  hill  what- 

*  This  value  of  ha   is  the  value  of  x    (for  the  given  constant  value  of  y) 
when  2  =  0,  as  found  from  equation  (1). 


102 


CALCULUS. 


ever  can  be  found  in  the  same  way.  Find  by  a  first  integration, 
the  area  ^4  of  a  plane  section  of  the  hill  parallel  to  the  xz  plane 
and  at  a  distance  y  therefrom,  and  then  integrate  A  •  dy  to 
get  the  required  volume;  or  find  by  a  first  integration  the  area 
A'  of  a  plane  section  of  the  hill  parallel  to  the  yz  plane  and  at  a 
distance  x  therefrom,  and  then  integrate  A'  *  dx  to  get  the 
required  volume. 

The  prismoid  formula. — A  prismoid  is  a  solid  with  parallel  top 
and  bottom  faces  with  sides  generated  by  straight  lines.  Thus  a 
cylinder  is  a  prismoid,  any  wedge  even  if  its  base  is  a  polygon  or 
curve  is  a  prismoid,  Fig.  47a  represents  a  prismoid. 

The  volume  of  a  prismoid  is  given  by  the  formula 


T  +  ^M  +  B 
6 


Xh 


(6) 


where  T  is  the  area  of  the  top  face,  B  is  the  area  of  the  bottom 
face,  M  is  the  area  of  the  middle  section  (which  is  parallel  to 
top  and  bottom  faces)  and  h  is  the  altitude  of  the  prismoid. 

Equation  (6)  also  gives  the  volume  of  any  figure  bounded  by  a 
second  degree  surface    ss    in  Fig.  476  between  parallel  top  and 


Fig.  47a. 


Fig.  476. 


bottom  faces,  and  equation  (6)  can  be  used  to  give  a  very  close 
approximation  to  the  volume  of  any  solid  between  parallel  ends 
with  sides  smoothly  curved  from  end  to  end. 


PARTIAL  DIFFERENTIATION  AND  INTEGRATION.       103 


PROBLEMS. 

1.  Integrate  equations  (4)  and  (5),  above,  and  find  the  volume 
of  one  quarter  of  a  hemispherical  hill  as  shown  in  Fig.  41a.  Verify 
the  result  from  the  formula:  volume  of  a  sphere  equals  ^tt  times 
its  diameter  cubed. 

2.  Find  the  volume  of  the  entire  wedge  (20  inches  long)  in  Fig. 
p2,  the  wedge  being  8  inches  wide  in  a  direction  perpendicular  to 
the  plane  of  the  paper. 

Ans.     800  cubic  inches. 

Note. — The  volume  of  a  wedge  is,  of  course,  easily  found  by  using  the  prin- 
ciples of  elementary  geometry.  It  is 
intended,  however,  that  this  problem  be 
solved  by  integration  as  follows:  Let  v 
be  the  volume  of  the  shaded  portion  of 
the  wedge  as  shown  in  Fig,  p2.  Then 
t;  is  a  function  of  x,  and  the  increment 
of  V  due  to  an  increment  of  a:  is  the 
volume  of  the  thin  slab  shown  in  Fig.  ^2. 

The  area  of  the  face  of  the  slab  is    — 

X  10  inches  X  8  inches  =  4a:  square  inches. 
Therefore  the  volume  of  the  slab  is  4a;-dx 
cubic  inches.    That  is, 

dv  =  Ax'dx 


^x-axU 


Fig.  2)2. 


and  the  desired  volume  is  found  by  integrating  this  expression  from  ic  =  0  to 
aj  =  20  inches. 

3.  Find  the  volume  of  the  shaded  portion  of  the  wedge  in 


30  inches 

Fig.  p3. 


jj i0_inche8 

Fig.  p4. 


104 


CALCULUS. 


Fig.  p3,  the  wedge  being  6  inches  wide  in  a  direction  perpendicular 
to  the  plane  of  the  paper. 
Ans.     92 L6  cubic  inches. 

4.  Find  the  volume  of  the  frustum  of  a  cone  which  is  shown  in 
Fig.  p4:. 

Ans.    229.2  cubic  inches. 

5.  Find  the  volume  of  the  paraboloid  of  revolution  which  is 
shown  in  Fig.  p5. 

Ans.     1060.7  cubic  inches. 

Note. — ^The  equation  of  the  paraboloid  is  y  =  px',  where  the  axis  of 
revolution  is  the  ^/-axis  of  reference;  and  the  value  of  p  is  determined  from 
the  condition  that  x  =  7}4  inches  when  y  =  12  inches. 


'  7  inches 
f 


5  inches 


Fig.  p5. 


Fig.  p6. 


6.  Find  the  volume  of  the  shaded  portion  of  the  paraboloid  in 
Fig.  p6. 

Ans.    699.8  cubic  inches. 

7.  Find  the  volume  of  the  spherical  segment  which  is  shown  in 
Fig.  p7. 

Ans.    276.79  cubic  inches. 

8.  Find  the  volume  of  the  segment  of  a  sphere  which  is  shown 
in  Fig.  pS. 

Ans.     119.71  cubic  inches. 
'  9.  Find  the  volume  between  the     xy     plane  and  the  plane 
z  =  a  -{-hx  +  cy  and  inside  of  the  cylinder  {x  —  ey-\-{y—fY  =  g^ 
when  a  =  10  inches,    6  =  3,   c  =  2,   e  =  8  inches,  /  =  9  inches 
and  gr  =  5  inches. 

Ans.    4085.7  cubic  inches. 


PARTIAL  DIFFERENTIATION  AND  INTEGRATION.       105 

Note. — Let  A-dy  be  the  expression  which  must  be  integrated  with  respect 
to  y  to  give  the  desired  volume.  Then  A  is  obtained  by  integrating  z  •  dx 
between  Umits  which  are  the  two  values  of  x  found  from  (x  —  e)^  +  (?/  —  ff 
=  g^  for  the  given  value  of  y,  and  then  A-dy  is  integrated  between  the 
limits  y  ^  f  -\-  g  and  y  =  f  —  g. 


I 


Is 


Fig.  p7. 


finches 
Pinches^  ^ 


'^ 


Fig.  p8. 


I 
I 

II 


li 
1^. 


-^^ 


10.  Find  the  total  volume  which  is  common  to  the  two  cylinders: 
1/2  _|_  2-2  _  25  and  x^  +  y^  =  25,  everything  being  expressed  in 
inches. 

Ans.     666.7  cubic  inches. 

Note. — One  should  be  able  to  determine  the  proper  limits  for  each  integra- 
tion in  this  problem  if  one  understands  problem  9. 

11.  The  inside  dimensions  of  a  barrel  are  20  inches  in  diameter 
at  each  end,  23  inches  in  diameter  at  the  middle  and  31  inches  long. 
What  is  the  capacity  of  the  barrel  in  gallons? 

Ans.    51.3  gallons. 

69.  Area  of  the  surface  of  a  hill.  Another  example  of  partial 
integration. — It  is  desired  to  find  the  area  of  the  portion  caCA 
of  the  surface  of  the  hill  shown  in  Fig.  48  when  the  equation  of 
the  surface  of  the  hill  is  given;  this  equation,  of  course,  expresses 
the  height  z  of  the  hill  at  a  point  as  a  function  of  x  and  y  as 
explained  in  Art.  68.     Let    S    be  the  area  of    caCA.    Then  the 


106 


CALCULUS. 


normal  to  dt 


increment  of    S    due  to  an  infinitesimal  increment  of    y    {=  Bb 

in  the  figure)  is  the  area  dS  of 
the  narrow  strip  ca.  To  find  the 
area  of  this  narrow  strip  let  s 
be  the  area  of  the  portion  from 
c  to  ds  in  the  figure,  then  s  is  a 
function  of  x,  and  the  increment 
of  s  due  to  an  infinitesimal  in- 
crement of  X  is  the  small  ele- 
ment of  area  ds  which  caps  the 
prism  dx  .  dy.  This  small  ele- 
ment of  area  is  sensibly  coinci- 
dent with  the  tangent  plane  at 
ds,  and  therefore  the  normal  to 
Fig.  48.  ds  makes  an  angle  7    with  the 

z-axis  whose  tangent  is  equal  to 


^/(i^(i)' 


which  is  the  resultant  slope  of  the  hill  at    ds, 


according  to  Art.  64. 

Now  the  area  of  the  base  of  the  prism,  dx 
of  the  area  ds,   and  consequently  we  have ; 

dx  '  dy  —  cos  7  •  ds 
dx  •  dy 


dy,  is  the  projection 


or 


ds  = 


But  if 
we  find: 


cos  7 


(1) 


<-^'4(MFW)' 


COS  7  = 


^Rl)■-(sJ 

so  that  equation  (1)  becomes: 


(2) 


'^  =  ^'  +  {'iy+{fJ-'^-'y         (2) 


.Jjl^ 


PARTIAL  DIFFERENTIATION  AND  INTEGRATION.       107 


10  inches 


Now  the  expression  under  the  radical  is  a  function  of  x  and  y 
which  may  be  found  by  differentiating  the  equation  of  the  surface 
of  the  hill  (which  is  given);  and  the  total  area  of  the  strip  ca 
(which  is  equal  to  dS)  is  the  integral  of  (3)  from  x  =  0  to  x  =  ha* 
In  this  integration  y  and  dy  are  constants.  The  value  of  dS 
so  found  is  the  required  infinitesimal  increment  of  the  area  caCA 
due  to  an  infinitesimal  increment  of  y,  and  the  value  of  the  area 
caCA  is  found  by  integrating  this  expression  for  dS  between  the 
limits  y  =  0  and  y  =  y  (meaning  any  value  of  y).  If  the 
entire  area  of  the  hill  in  Fig.  48  is  to  be  found,  the  expression  for 
dS  must  be  integrated  from  2/  =  0  to  y  =  the  intercept  of  the 
hill  on  the  ^/-axis.f 

PROBLEMS. 

1 .  The  hill  shown  in  Fig.  48  is  a  quarter  of  a  hemisphere.    Find 
the  area  of  its  surface  and  verify 
your   result   from   the    formula: 
area  of  a  sphere  =  47r  times  its 
radius  squared. 

2.  Find  the  area  of  the  curved 
surface  of  the  cone  shown  in 
Fig.  p2, 

Ans.     135.8  square  inches. 

Note. — Let    a    be  the  area  of  the       i            jp  ly-i 

shaded  portion  of  th  cone  in  Fig.  p2.  "  *" 

Then  a  is  a  function  of  x;  and  the  in-  Yig.  v2. 
crement  of  a  due  to  an  increment  of  x 

is  the  area  of  the  narrow  strip  shown  in  the  figure.  The  width  of  this  strip 

(parallel  to  the  slant  height  of  the  cone)  is    ^^^  "^  ^^  X  dx,   and  the  length 

of  the  strip  (cu-cumference  of  the  cone)  is  ~.XS  X  tt  inches.    Therefore  we 

get: 

da  =  OM^irX'dx 

*  The  value  of  6a  is  found  from  the  given  equation  of  the  surface  of  the  hill. 
It  is  the  value  of  x  for  the  given  constant  value  of  y  when  2  =  0. 

t  This  intercept  is  found  from  the  equation  of  the  surface  of  the  hill  by  plac- 
ing X  and  z  both  equal  to  zero  and  solving  for  y. 


108 


CALCULUS. 


and  the  required  area  is  found  by  integrating  this  expression  from   a;  =  0  to 
X  =  10  inches. 

3.  Find  the  area  of  the  curved  surface  of  the  frustum  of  the 
cone  in  Fig.  p3. 

Ans.     144.38  square  inches. 

4.  Find  the  area  of  the  part  of  the  surface  of  the  cylinder 


.^^' 


^>.. 


j^ JUf.Jnche8 ^ 

Ilg.p3. 


^1 


4  inches 
Pinches, 


m 


1^ 


I 


Fig.  p5. 


a;2  -f  ^2  _  100  which  is  inside  of  the  cylinder  x^  -\-  y^  =  100,  the 
radii  of  the  cylinders  being  measured  in  inches. 
Ans.    800  square  inches. 

5.  Find  the  area  of  the  convex  surface  of  the  segment  of  a 
sphere  which  is  shown  in  Fig.  p5. 

Ans.     989.5  square  inches. 

6.  Find  the  area  of  a  zone  on  the  earth's  surface  between  40° 
and  80°  north  latitude,  taking  the  earth  as  a  sphere  4,000  miles 
in  radius. 

Ans.    34,500,000  square  miles. 


CHAPTER  V. 
MISCELLANEOUS  APPLICATIONS. 

70.  Influence  of  errors  of  observation  upon  a  result.  When 
one  measures  a  thing  with  great  care  repeatedly  the  successive 
measurements  never  agree  with  each  other,  and  it  is  easy  to 
determine  what  is  called  the  probable  error*  of  the  measurement 
by  considering  the  discrepancies  in  a  set  of  observed  values. 
Thus  the  length  of  a  bar  might  be  found  by  repeated  measure- 
ment to  be  103.625  centimeters  with  a  probable  error  of,  say, 
db  0.0036  centimeter;  the  mass  of  a  body,  as  determined  by  re- 
peated weighing  on  a  balance,  might  be  67.2382  grams  with  a 
probable  error  of,  say,  ±  0.0006  gram. 

In  many  cases,  however,  it  is  not  the  directly  measured 
quantities  that  are  important  but  a  result  which  is  calculated 
therefrom;  and  it  is  often  desirable  to  calculate  the  probable 
error  in  such  a  result  when  the  probable  errors  in  the  individual 
measurements  are  known.  For  the  sake  of  simplicity  we  will 
discuss  a  special  case,  as  follows: 

A  steel  ball  is  weighed  on  a  balance  and  its  mass  is  found  to  be 
M  grams  with  a  probable  error  of  ±  m  grams,  and  the  diameter 
of  the  ball  is  measured  and  found  to  be  L  centimeters  with  a 
probable  error  of  ±  I  centimeters.  The  density  of  the  steel 
as  calculated  from  the  measured  values  of  M  and  L  is: 

^  =  &  (^) 

and  it  is  required  to  find  the  probable  error  in  D.     The  probable 

*See  Franklin,  Crawford  and  MacNutt's  Praclicol  Physics,  Vol.  I,  pages 
1-9  for  a  very  simple  discussion  of  errors  of  observation.  For  more  complete 
discussion  see  Merriman's  Least  Squares,  John  Wiley  &  Sons,  New  York.  See 
also  Palmer's  Theory  of  Measurements,  McGraw-Hill  Co.,  New  York,  1912. 

109 


110  CALCULUS. 

errors  m  and  I  are  usually  very  small  so  that  it  is  sufficiently 
accurate  to  calculate  ADm  (the  probable  error  in  D  due  to 
=t  m)  and  ADi  (the  probable  error  in  D  due  to  ±  I)  by  the 
formulas: 

AI>«  =  l^-m  (2) 

and 

AD,  =  gj  (3) 

Let  AD  be  the  probable  error  in  the  result  due  to  m  and  I 
together.  Then  AD  is  not  equal  to  the  sum  AD;;^  +  AD^,  but 
according  to  the  theory  of  probability  we  have: 

AD  =  V(ADJ2  +  (ADO^  (4) 

PROBLEMS. 

1 .  The  scale  of  an  alternating  current  ammeter  reads  in  degrees, 
and  the  value  of  the  current  is  i  =  k^,  where  fc  is  a  constant 
and  0  is  the  deflection  in  degrees  due  to  a  current  of  i  amperes. 
The  error  of  reading  is  probably  ±  i^  degree.  What  is  the 
probable  error  in  the  result,  t,  in  fractions  of  an  ampere  (a)  when 
the  deflection  is  in  the  neighborhood  of  200°  and  (6)  when  the 
deflection  is  in  the  neighborhood  of  20°?  Take  the  value  of  k 
to  be  such  that  10  amperes  gives  100°  deflection.  Ans:  (a) 
=t  0.0089  ampere,  (6)  ±  0.028  ampere. 

2.  What  is  the  percentage  error  (probable)  in  the  value  of  i 
under  conditions  (a)  and  (6)  in  problem  1?  Ans.  (a)  =t  0.06  per 
cent.,  (6)  ±  0.6  per  cent. 

Note.    The  percentage  error  is  —  X  100. 

3.  The  diameter  of  a  steel  sphere  is  repeatedly  measured  and 
found  to  be  0.8762  inch  with  a  probable  error  of  ±  0.0016  inch. 
Calculate  the  volume  of  the  sphere  and  calculate  the  probable 
error  in  the  calculated  volume.  Ans.  0.3525  cubic  inch 
=b  0.0019  cubic  inch. 


MISCELLANEOUS  APPLICATIONS. 


Ill 


Note.     It  is  interesting  to  note  that  the  percentage  error  in  the  volume  i3 
three  times  as  great  as  the  percentage  error  in  the  measured  diameter,  that  is 


0.0019  . 


0.3525 


is  three  times  as  great  as 


0.0016 
0.8762 ' 


4.  Show  that  the  percentage  error  of  x^  is  twice  as  great  as 
the  percentage  error  of  x,  when  a;  is  an  observed  quantity  and 
x^  is  a  calculated  result. 

5.  By  measurement  the  length  of  a  rectangle  is  25  inches 
with  a  probable  error  of  =^  0.02  inch,  and  the  width  of  the  rec- 
tangle is  10  inches  with  a  probable  error  of  ±  0.015  inch.  What 
is  the  probable  error  in  the  calculated  area  of  250  square  inches? 
Ans.     ±  0.421  square  inch. 

6.  The  diameter  of  a  steel  ball  is  2.542  centimeters  ±  0.001 
centimeter,  and  the  mass  is  116.25  grams  ±  0.02  gram.  What  is 
the  density  and  what  is  the  probable  error  in  the  density?  Ans. 
7.781  grams  per  cubic  centimeter  =t  0.0168  gram  per  cubic 
centimeter. 

71.  Maximum  and  minimum  values.  It  is  evident  that  a 
growing  thing  reaches  its  maximum  size  when  it  stops  growing; 
but  a  thing  may  stop  growing  for  a  while  and  then  go  on  growing 


Fig.  49. 

.         .  dv  .         ... 

A  maximum;  -^  is  positive  when  x  is 

less  than  a,   and  negative  when   x  is 
greater  than  a. 


Fig.  50. 
Not  a  maximum;  -^  is  positive  when 


X    is  less  than    a, 
greater  than  a. 


also  when    x    is 


again.  Also  it  is  evident  that  a  decreasing  thing  reaches  its 
minimum  size  when  it  stops  decreasing;  but  a  thing  may  stop 
decreasing  for  a  while  and  then  go  on  decreasing  again.     Thus 


112 


CALCULUS. 


■^    is  equal  to  zero  when    x  =  a    in  all  four  of  the  following 

figures,  49,  50,  51  and  52,  but    6    is  a  maximum  value  of   y   in 
Fig.  49  only,  and   6   is  a  minimum  value  of   y   in  Fig.  51  only. 


y-ojdB 


I        x-axi8 
. —  a ->l 

Fig.  51. 

dti 
A  minimum;  -^  is  negative  when  x 

is  less  tlffen   a,   and  positive  when    x 
is  greater  than  a. 


y-€ixi8 


Fig.  52. 

dv 
Not  a  minimum;    -^    is  negative 

when    X    is  less  than    a,    also  when 
X  is  greater  than  a. 


To  determine  the  values  of  x  for  which  a  given  function  of  x 
is  a  maximum  or  a  minimum,  differentiate  the  function,  place  its 
derivative  "equal-  to  zero,-and  'solve  for  x.  Then  for  each  value  a 
of  X  so  found,  determine  the  algebraic  sign  of  the  derivative  for  a 
value  of  X  slightly  less  than  a  and  also  for  a  value  of  x  slightly 
greater  than  a.     Interpret  the  results  by  referring  to  Figs.  49  to  52* 


PROBLEMS. 

In  the  first  five  problems,  following,  find  for  what  values  of  x 
is  2/  a  maximum  or  minimum  and  compute  the  maximum  or 
minimum  value  in^each  case.  • 

1.  2/  =  a:'' —  6x2  +  9x  -f  3.  ^^s.  x  =  1)  y  =  7,  a  maxi- 
mum:   X  =  3;  2/  =  3,   a  minimum. 

^  x^  +  4x  +  44 

2.  2/  = 


Ans. 


a^  =  8;  2/  =  I, 


a  mimmum: 


x2  +  8x  +  40 
X  =  —  6;  2/  =  2,   a  maximum. 

*  A  slightly  easier  method  is  to  consider  the  values  of  the  second  derivative 
of  the  given  function  but  the  above  method  is  sufficient. 


MISCELLANEOUS  APPLICATIONS.  113 

3.  y  =  (x  —  4)3  {x  +  ly.  Ans.  x  =  —  1;  y  =  Oj  a  maxi- 
mum :  X  =  1;  y  =  —  108,   a  minimum. 

4.  y  =  9e2=^  +  25e~2^.     Ans.  a:  =  log  Vf;  2/  =  30,  a  minimum 

5.  1/  =  3  sin  a;  +  4  cos  a;.  Ans.  a;  =  tan~^f  in  the  first  quad- 
rant; y  =  5,  a  maximum:  x  =  tan-^J  in  the  third  quadrant; 
y  =  —  6,    a  minimum. 

6.  Divide  the  number  20  into  two  parts  such  that  the  product 
of  one  part  by  the  square  of  the  other  part  shall  be  a  maximum. 
Ans.  13.33  and  6.67. 

7.  What  number  added  to  one  half  the  square  of  its  reciprocal 
gives  the  least  possible  sum.     Ans.  1. 

8.  The  cost  of  an  electrical  power  transmission  line  is  approxi- 
mately proportional  to  the  weight  of  copper  used  because  the 
cost  of  poles  and  the  cost  of  erection  can  be  expressed  with  a  fair 
degree  of  accuracy  as  an  addition  of  so  many  cents  per  pound  on 
the  cost  of  copper.  Therefore  the  annual  money  cost  represented 
by  interest  on  investment  and  depreciation,  being  reckoned  as 
so  many  per  cent,  of  first  cost,  is  proportional  to  the  weight  of 
copper  used,  or  it  is  equal  to  kw  where  A;  is  a  constant  and  w 
is  the  weight  of  copper  used. 

The  amount  of  power  lost  in  a  transmission  line  is  halved  if 
the  amount  of  copper  is  doubled.  Therefore  the  annual  money 
loss  due  to  loss  of  energy  in  the  line  is  universely  proportional 

to  the  weight  of  copper,  or  it  is  equal  to   —   where  m  is  a  con- 
stant. 

Therefore  the  total  annual  money  loss  is   kw  H .     Find  the 

w 

value  of   w    (expressed  in  terms  of   k   and   w)  which  will  make 
this  total  annual  money  loss  a  minimum.    Ans.  w  =  ^llH 

Note.  This  matter  is  discussed  in  electrical  engineering  books  under  the 
head  of  Kelvin's  law.  See  Franklin's  Electric  Lighting^  Art.  26,  pages  66-70. 
The  Macmillan  Co.,  New  York,  1912.' 

8.  It  is  desired  to  construct  of  sheet  metal  a  cylindrical  gallon 


114  CALCULUS. 

measure  without  a  cover.     Find  what  its  depth  must  be  in 
order  that  the  amount  of  sheet  metal  used  shall  be  a  minimum. 


Ans.  y  =  jj?^  (=  4,19)   inches. 


Note.  The  amount  of  sheet  metal  used  will  be  a  minimum  when  the 
surface  area  S  of  the  vessel  is  a  minimum;  but  S  =  2irxy  +  ttx^,  where  x 
is  the  radius  of  the  base  of  the  cylinder  and  y  is  its  depth.  Since  the  measure 
is  to  contain  1  gallon  (=231  cubic  inches),  irx^y  =231;  and,  substituting  for 
z  its  value  in  terms  of  y,  we  have: 

^  =  2>/23i^  +  ??i 

The  value  of   y   which  makes  this  expreasion  a  minimum  is  to  be  obtained. 

10.  Determine  the  altitude  of  the  cylinder  of  greatest  convex 
surface  that  can  be  inscribed  in  a  sphere  whose  radius  is  10  inches. 
Ans.  14.142  inches. 

11.  Determine  the  altitude  of  the  cylinder  of  greatest  volume 
that  can  be  inscribed  in  a  sphere  whose  radius  is  10  inches. 
Ans.  11.547  inches. 

12.  Determine  the  volume  of  the  smallest  cone  which  can  be 
circumscribed  about  a  sphere  whose  radius  is  10  inches.     Ans. 

— ^—    (=  838.1)  cubic  inches. 
o 

13.  A  man  who  can  row  at  a  speed  of  4  miles  per  hour  and 

run  at  a  speed  of  6  miles  per 
hour    wishes    to    reach    the 

gi     \  point     p     from  a  boat  at   h 

%^->.^  ^\  as  shown  in  Fig.  pl3  in  the 

^1      \     \^^  least  possible  time.    Find  the 

T^ch — ^        ^     V.  '  beach      distance    ap     that  the  man 

t;,.  '"'-t*  must  run  on  the  beach.    Ans. 

rig.  pl3. 

1.056  miles. 

14.  The  strength  of  a  rectangular  beam  is  proportional  to  its 
breadth  and  to  the  square  of  its  depth.  Find  the  breadth  and 
depth  of  the  strongest  beam  which  can  be  cut  from  a  cylindrical 


'N 


MISCELLANEOUS  APPLICATIONS.  115 

log  whose  radius  is  18  inches.  Ans.  Breadth,  20.8  inches;  depth, 
29.4  inches. 

15.  For  what  angle  of  deflection  does  the  old-fashioned  tangent 
galvanometer  give  the  greatest  percentage  accuracy  in  the  value 
of  the  current?     Ans.  45°. 

Note.    The  equation  of  the  tangent  galvanometer  is: 

i  =  k  tan  6 

where  i  is  the  current  in  amperes,  6  is  the  observed  angle  of  deflection,  and 
fc  is  a  constant.  Let  dd  represent  the  probable  error  of  the  observed  value 
of    0,   then  the  probable  error  in  the  value  of  i  is: 

,.       k.de 
di  = 


COS2  d 

and  the  percentage  error  in  the  value  of  i  is: 

_di  _        dd 
^  ~  i  ""  sin  0  COS  9 

in  which  dd  being  the  probable  error  of  6  is  to  be  treated  as  a  constant. 

A  general  discussion  of  this  important  matter  of  precision  as  dependent 
upon  the  choice  of  magnitudes  of  the  quantities  to  be  observed  is  given  in 
Chapter  XII  of  Palmer's  Theory  of  MeasurementSy  McGraw-Hill  Co.,  New 
York,  1912. 

72.  The  problem  of  the  bent  beam.     The  formulas  used  in 

connection  with  a  bent  beam  are  based  on  Hookers  law,  and  the 

F       sectional  area  a  p  p        sectional  area  8  p 

original  length  L     m!  original  length  L        y 

increase  of  length  I --^  decrease  of  length  l-^ 

Fig.  53.  Fig.  54. 

derivation  of  two  of  the  simpler  formulas  furnishes  a  good 

example  of  the  use  of  calculus. 

Hookers  law.     Figure   53   represents   a   rod   subjected   to   a 

tp 

stretching  force   F.     Experiment  shows  that   —   is  proportional 

s 


116  CALCULUS. 

to    y   (Hookers  law).    Therefore  we  may  write 


L 


or 


a  L 


F  =  E-^  (1) 


in  which  ^  is  a  proportionality  factor  and  it  is  called  the  stretch 
modulus  of  the  substance  of  which  the  rod  is  made.  Equation 
(1)  also  gives  the  compressing  force  F  required  to  produce  a 
slight  shortening  of  the  rod  as  represented  in  Fig.  54. 

When  a  beam  is  bent  the  filaments  on  one  side  of  the  beam  are 
lengthened,  and  the  filaments  on  the  other  side  are  shortened. 
If  we  wish  to  determine  the  degree  of  lengthening  or  shortening 
it  is  necessary  to  consider  a  very  short  portion  of  the  beam. 
But  a  very  short  portion  of  a  bent  beam  has  the  same  curvature 
as  the  osculating  circle  as  shown  in  Fig.  55,  and  the  easiest  way 

^^  mediaa  Une 


very  short  portion 
beam 


Fig.  55.  Fig.  66. 

to  think  of  the  very  short  portion  of  the  bent  beam  is  to  think  of  it 
as  a  portion  of  a  long  beam  which  is  actually  bent  into  the  arc  of  a 
circle.  Thus  Fig.  56  represents  a  beam  bent  into  the  arc  of  a 
circle. 

A  certain  filament  in  the  bent  beam  is  unchanged  in  length. 


MISCELLANEOUS  APPLICATIONS. 


117 


pulU 


This  filament  is  located  in  what  is  called  the  median  line  (or 
plane)  of  the  beam.  The  median  plane  of  a  beam  of  rectangular 
section  is  the  middle  plane  of  the  beam. 

Let  us  consider  the  forces  which  act  across  a  section  qq  of  a 
bent  beam  as  shown  in  Fig.  57.  Considering  these  forces  as 
acting  on  A5,  they  are  a  set  of 
pulls  on  A  and  a  set  of  pushes 
on  B,  and  together  they  con- 
stitute a  turning  force  or  torque, 
T,  about  an  axis  0  perpen- 
dicular to  the  plane  of  the  paper. 
It  is  the  object  of  this  discussion 
to  derive  an  expression  for  T 
in  terms  of  6,  D,  R  and  Ej 
where    h    is  the  breadth  of  the 

beam,  D  is  the  depth  of  the  beam,  R  is  the  radius  of  curvature 
of  the  median  line  of  the  beam,  and  E  is  the  stretch  modulus 
of  the  material  of  which  the  beam  is  made. 

Let  us  consider  the  portion  of  the  beam  which  lies  between 
the  radii   RR  in  Fig.  56.     This  portion  of  the  beam  is  shown  to 


I 

I       median  line 

jIL.  — .TP . 


\x 


length  MR+x)^ 
length  s=Re 

length  =(R-jc;d 


Fig.  58. 

a  larger  scale  in  Fig.  58.     The  original  length  of  every  filament 
of  this  portion  of  the  beam  is   Rd.     Consider  the  upper  filament 


118 


CALCULUS. 


Jf  of  the  beam.  The  stretched  length  of  this  filament  is 
{R  +  x)dj  because  //  is  a  circular  arc  of  radius  (R  +  x)  and 
it  subtends  the  angle  6.  Therefore  the  filament  }}  has  been 
increased  in  length  by  the  amount  xB  by  the  bending.     Similarly 

it  may  be  shown  that  the  lower 


stretched 


_      5    /'Stretch 


eompresaed- 

•ectional  viewjof  Jtean 

Fig.  59. 


filament  /'/'  in  Fig.  58  has  been 
shortened  by  the  amount  xd  by 
the  bending. 

Figure  59  is  a  sectional  view 
of  the  beam  with  its  depth  arbi- 
trarily increased  by  added  layers 
of  material  on  top  and  bottom, 
and  we  wish  to  find  the  incre- 
ment of  T  due  to  this  assumed 
increase  of  depth  of  the  beam. 
The  stretching  force  in  the  top 

layer  can  be  found  from  equation  (1)  by  substituting  xd  for   Z, 

h .  dx  for  s  and  RO  for  L,   which  gives 

F  =  -r^'X'dx 

this  stretching  force  is  a  pull  at  A  in  Fig.  57,  and  its  torque 
action  about  the  axis  00  in  Fig.  59  is  counter-clock  wise  as 
seen  in  Fig.  57,  and  it  is  equal  to 

Fx  =  ^'xHx 

The  compressing  force  in  the  bottom  layer  is  a  push  at  B 
in  Fig.  57,  and  its  torque  action  about  the  axis  00  is  also  counter- 
clock  wise  as  seen  in  Fig.  57,  and  it  is  equal  to 


Fx  = 


Eh 
R 


x^'dx 


Therefore  the  total  torque  action  due  to  the  two  added  layers  is: 
dT  =  ?|^-x2.daj  (2) 


MISCELLANEOUS  APPLICATIONS. 


119 


Whence,  by  integrating,  we  have: 
2  Ehx^ 


T  = 


3     R 


+  a  constant 


(3) 


But  it  is  evident  that  T  =  0  when  x  =  0,  because  a  beam  of 
zero  depth  would  of  course  have  no  stiffness  at  aU.  Therefore 
the  constant  of  integration  in  equation  (3)  is  zero,  and  equation 
(3)  becomes: 


(4) 


It  is  somewhat  simpler  to  express  T  in  terms  of  the  depth  D 
of  the  beam  {=  2x).  Therefore  substituting  x  =  ^  inequation 
(4),  we  have: 


12    B 


(5) 


This  equation  applies  to  a  beam  bent  in  any  way  whatever,    T 

being    the    torque    action 

across  a  particular  section 

of  the  beam,  and   R  the 

radius  of  curvature  of  the 

median  line  at  that  point. 

Thus  Fig.  60  shows  a  beam 

with  one  end  fixed  in  a  wall 

and  carrying  a  weight    W 

at  the  other  end.     In  this 

case  the  bending  torque  at 

p    is  equal  to    W(a  —  x)j 

and  the  torque  action  across 

the  section  of  the  beam  at 

p  is  equal  and  opposite  to 

W{a  —  x)     Therefore,  ignoring  algebraic  signs,  we  have 


Fig.  60. 


T  =  W(a  -  x) 


(6) 


120  CALCULUS. 

Substituting  this  value  for   T  in  equation  (5)  and  solving  for  R 

we  have: 

_  1       EhD^ 

^~T2W(a-x)  ^^^ 

or 

R  =  -^  (8) 

a  —  X 

where 

c  =  -i.^^  (9) 

Let  it  be  required  to  find  the  equation  of  the  curve  formed  by 
the  median  line  of  the  beam  in  Fig.  60.  According  to  Art.  49 
the  radius  of  curvature  of  any  curve  at  a  point  is: 


n.hM 


If  the  bending  of  the  beam  in  Fig.  60  is  very  slight  the  value  of 
-p  will  be  everywhere  very  small  so  that,  as  a  first  approxima- 
tion, we  may  neglect  (~f)  in  the  numerator  of  equation  (10)  where 

it  is  added  to  the  much  larger  constant  quantity  1.  Therefore 
from  equations  (8)  and  (10)  we  have 

i  =      ^ 

(Py      a  —  X 

dx" 
or 

d^y  _  a      x 

dx^       c      c 
whence,  by  integrating,*  we  have 


(11) 


(12) 


dy      ax       x^  ,  .     . 

;p  = —  +  Si  constant 

uX  C  JiiC 


(13) 


*  By  recognizing  the  function  of  which is  the  derivative. 


MISCELLANEOUS  APPLICATIONS. 


121 


But    ~    is  actually*  equal  to  zero  when    a;  =  0    in  Fig.  60, 
that  the  constant  of  integration  in  equation  (13)  is  zero,  or: 


so 


dy 
dx 


ax 
c 


2c 


Integrating  this  expression,  we  have: 


y  = 


ax2 

2c  ~  6c 


+  a  constant 


(14) 


(15) 


But    y  =  0    when    x  =  0    in  Fig.  60,  so  that  the  constant  of 
integration  in  equation  (15)  is  zero,  and  equation  (15)  becomes: 


y 


ax" 
2c 


6c 


(16) 


in  which  c  the  constant  defined  by  equation  (9). 

73.  Average  value  of  a  function.  Let  y  he  sl  function  of  x 
as  represented  by  the  curve 
cc  in  Fig.  61.  The  average 
value  of  y  between  x  =  a 
and  X  =  h  is  the  shaded  area 
in  Fig.  61  divided  by  the  base 
{b  —  a).  This  gives  the  height 
of  a  rectangle  of  base  {h  —  a) 
the  area  of  the  rectangle  being 
equal  to  the  shaded  area  in 
the  figure.     From  this  definition  of  average  value  we  have: 

{average  value  of  y 
between  x  =  a  and  x 


Fig.  61. 


=i]^^a£y- 


dx       (1) 


PROBLEMS. 


1.  The  velocity  of  a  falling  body  is  v  =  gt.     Find  the  average 
value  of  V  from  i  =  0  to  t  =  10.     Ans.  5g. 
*  Not  approximately. 


122 


CALCULUS. 


2.  The  velocity  of  a  falling  body,  starting  with  an  initial 
velocity  w,  is  v  =  u  -\-  gt.  Find  the  average  value  of  v  from 
t  =  5  to  <  =  10.     Ans.  u  +  7.5  g. 

3.  Find  the  average  value  of  y  —  sinx  from  a;  =  0  to  a;  =  tt. 
The  answer  is  shown  in  Fig.  p3. 


y=8inx 


Fig.  p3. 

4.  Find  the  average  value  of  e  =  E  sin  (at  during  a  whole 
cycle  from   cot  =  0  to  w^  =  27r.     Ans.  Zero. 

5.  Find  the  average  value  oi  y  =  sin^  x  during  a  half  cycle 
(x  =  0  to  X  =  tt)  and  during  a  whole  cycle  (x  =  0  to  x  =  27r). 
The  answer  is  shown  in  Fig.  p5. 


Fig.  p5. 

6.  Find  the  average  value  oi   y  =  sin  a;  sin  {x  —  0)    during  a 
whole  cycle   {x  —  0  to  x  —  2t).      Ans.  J^  cos  d. 


MISCELLANEOUS  APPLICATIONS.  123 

CENTER  OF  GRAVITY. 

74.  Resultant  of  a  set  of  parallel  forces.     The  single  force   R 
in  Fig.  62  is  equivalent  to  the  three  parallel  forces  A,  B  and  C 

combined,  and  it  is  called  the    ^ ^ 

resultant  of    A,  B     and     C.    k X >j 

The  value  of   R    is  equal  to    it__a— ^  !  j 

A  -{-  B  -\-  C,  and  the  point  of  i^         j  j^.  ji 

apphcation  of  R  is  determined        ^       M  |b  1^1^ 

by    the    condition    that    the  } 

torque  action  of  R    about  any  Ijij 

arbitrarily    chosen     axis     0  X 

(perpendicular  to  the  plane  of  p-    g2. 

the  paper)  must  be  equal  to 

the  combined  torque  action  of  the  given  forces     A,  5,    and     C 

about  that  axis.     That  is: 

R^  A  +  B  +  C  (1) 

(2) 

A  +  B  +  C  ^^^ 

Definition  of  the  center  of  gravity  of  a  body.  Every  portion 
of  a  body  is  pulled  downwards  by  gravity,  and  the  forces  which 
thus  act  upon  the  various  parts  of  a  body  are  together  equivalent 
to  a  single  force  (their  resultant)  the  point  of  application  of  which 
is  called  the  center  of  gravity  of  the  body.  Thus  the  short  arrows 
///  in  Fig.  63  represent  the  forces  with  which  gravity  pulls  on  a 
body,  these  forces  are  together  equivalent  to  the  single  force  F, 
and  the  point  of  application  of  F  is  the  center  of  gravity,  C, 
of  the  body. 

The  force  with  which  gravity  pulls  on  a  body  is  proportional 
to  the  mass,  m,  of  the  body.  Therefore,  if  we  use  the  pull 
of  gravity  on  a  one-pound  body  as  our  unit  of  force,  the  pull 


and 

RX  =  Aa  +  Bh  +  Cc 

so  that 

■*, 

V  _Aa  +  Bh-{-Cc 

124 


CALCULUS. 


of  gravity  on  any  body  is  numerically  equal  to  the  mass  of  the 
body  in  pounds.     Thus  a  10-pound  body  would  be  pulled  by  10 

units  of  force.  This  unit 
of  force  is  called  a  pound. 
It  is  evident  therefore 
that  the  word  pound  has 
two  meanings;  it  signifies 
a  unit  of  mass  when  we 
speak  of  so  many  pounds 
of  sugar  or  coal,  and  it 
signifies  a  unit  of  force 
when  we  speak  of  so 
many  pounds  of  force. 
Consider  a  small  particle  of  the  body  in  Fig.  63  of  which  the 
mass  is  dm  pounds  and  of  which  the  abscissa  is  x  as  shown. 
The  pull  of  gravity  on  the  particle  is  dF  =  dm,  the  torque 
action  of  this  force  about  the  axis  0  (perpendicular  to  the  plane 
of  the  paper)  is  x-dm,  the  combined  torque  action  about  0 
of  all  the  forces  fff  may  be  expressed  as  the  integral  J  x-dm, 
and  this  total  torque  action  is  equal  to  the  torque  action  about 
0  of  the  resultant  F.     Therefore  we  have: 


Fig.  63. 


FX  =  Jx'dm 


(4) 


Now  the  force   F  is  equal  to  the  total  mass   M  of  the  body  as 
above  explained.     Therefore  from  equation  (4)  we  have: 


X  = 


fx'dm 


M 


(5) 


Similarly,  the  y  and  z  coordinates  of  the  center  of  gravity  are 
given  by  the  equations: 


F  = 


fy-dm 


M 


(6) 


MISCELLANEOUS  APPLICATIONS 


125 


and 


Z  = 


f  Z'dm 


M 


(7) 


Equations  (4)  and  (5)  have  exactly  the  same  significance  as 
equations  (2)  and  (3);  and  when  the  value  of  fx-dm  is  found 
for  the  whole  body,  then  the  distance  X  of  the  center  of  gravity 
from  the  origin  0  can  be  calculated  if  the  total  mass  of  the  body 
is  known. 

75.  Center  of  gravity  of  a  bent  rod.  It  is  desired  to  find  the 
coordinates  X  and  Y  of  the  center  of  gravity  of  a  rod  which  is 
represented  by  the  heavy  black  portion  of  the  parabola  which  is, 
shown  in  Fig.  64  and  whose  equation  is    y  =  kx^.     Let    h    be 


y-dxis 

1 

1 

I..  V  i 

\ 

\ 
\ 

/ 

r      ( 

1 

•-  1 

'^x. 

--■»< 

x-axifi 

t-"!-,- 

t 
1 

1 
/ 

/   1 

\ 
\ 

J    /    i 

/l>i 

N. 

,X     \           Ijc-ojcis 

^ JC  -^ 

Fig.  64.  Fig.  65. 

the  mass  of  the  rod  per  unit  of  length  (pounds  per  foot).     Then 


}i'd^  =  hA{dxY-\-  {dyf 

is  the  mass  in  pounds  of  the  portion  dfs  of  the  rod,  as  may  be 
understood  with  the  help  of  Fig.  65.  But  from  the  equation  of 
the  parabola  ,  y  —  fcx^,    we  have 


dy  —  2kx'dx. 


126  CALCULUS. 

Therefore  h  ^{dxY  +  {dyY  is  equal  to  ^  Vl  +  ^k'^x'^  •  dx  and 
this  expression  is  to  be  used  instead  of  dm  in  equation  (5)  of 
the  previous  article  so  that: 


£6 


Now  the  mass  of  the  rod  in  Fig.  64  is  the  length  of  the  rod 
multiplied  by  ^,  and  the  length  is  given  by  the  integral  of 
Vl  +  Wx^  •  dx  from  x  =  a  to  x  =  h.    That  is 


/'»x=6 


M  =  h  \      Vl  +  Wx'^'dx  (2) 

When   M   is  determined  by  this  equation  its  value  can  be  used 
in  equation  (1)  to  give  the  value  X. 

In  a  similar  manner  equation  (6)  of  Art.  74  gives,  for  the  bent 
rod  shown  in  Fig.  64: 


hh 


£=6 
xHl  -^Wx^'dx 
a 


Y — M (3) 

from  which  Y  can  be  calculated  when  M  is  known  Jsee  equation        ^ 
(2)]. 

76.  Center  of  gravity  of  a  solid  cone.  Consider  the  solid 
cone  shown  in  Fig.  66.  To  determine  the  location  of  the  center 
of  gravity  of  this  cone  equations  (5),  (6)  and  (7)  of  Art.  74  can 
be  used  as  in  the  case  of  a  bent  rod,  but  it  is  worth  while  to  give 
a  slightly  different  argument  as  follows:  Let  T  be  the  torque 
action  of  the  pull  of  gravity  on  the  cone,  T  being  reckoned  about 
the  axis  0  (perpendicular  to  the  plane  of  the  paper).  Then  T 
is  evidently  a  function  of  the  length  x  of  the  cone  {x  is  shown  in 
Fig.  67),  and  the  increment  of  T  due  to  an  increment  of  x  is: 

dT  =  X'dF  (1) 


J^ 


MISCELLANEOUS  APPLICATIONS. 


127 


where  dF  is  the  force  with  which  gravity  pulls  on  the  thin  slab 
of  material  in  Fig.  66,  the  radius  of  the  slab  being  y  and  its 
thickness  dx.     The  volume  of  this  slab  is  iry^  •  dx  and  its  mass  is 


Fig.  66. 


X'OxU 


Fig.  67. 


irDy^'dx  (=  dm)    where    D 
pounds  per  cubic  foot.     But 
of  the  half-angle  of  the  cone, 
so  that  equation  (1)  becomes 


is  the  density  of  the  material  in 
y  =  kx,  where  k  is  the  tangent 
Therefore  dm  =  dF  =  irkWx^'dx, 


whence 


dT  =  TrkWx^'dx 
T  =  — -i —  x^-\-  a.  constant 


(2) 
(3) 


but  it  is  evident  that  T  is  zero  when  x  =  0  because  when  x  =  0 
there  is  no  cone  at  all.  Therefore  the  constant  of  integration  in 
equation  (3)  is  equal  to  zero,  and  equation  (3)  becomes 


T=^-m.^ 


(4) 


Now  the  volume  of  the  cone  in  Fig.  66  is  a  function  of  x  whose 
increment  dV  is: 


dV  =  Try^'dx 


(5) 


128  CALCULUS, 

or,  using  y  =  A;a;  as  before,  we  have 

dV  =  wk^x^dx  (6) 

so  that 

V  =  iirk^x^  (7) 

the  constant  of  integration  being  zero  because  V  =  0  when 
X  =  0.  The  mass,  Af,  of  the  cone  is  equal  to  D7  pounds,  and 
the  total  pull  of  gravity  on  the  cone  is  therefore  F  =  DV  pounds. 
The  torque  action  of  this  force  about  the  axis  0  is 

FX  =  DVX  =  DX  times  JttAjV 

and  this  torque  action  is  equal  to  T.  Therefore,  substituting 
DX  times  iirk^x^  for  T  in  equation  (4)  and  solving  for  X, 
we  have: 

X  =  ix  (8) 

77.  Average  distance  of  the  particles  of  a  body  from  a  plane. 
A  new  view  of  center  of  gravity.  Imagine  a  body  to  be  made 
up  of  minute  particles  of  equal  mass.  Then  the  number  of 
particles  in  a  piece  of  the  body  would  be  proportional  to  the 
mass  of  the  piece,  that  is,  equal  to  A^  X  mass  of  piece,  where 
A^  is  a  constant.  Let  dm  be  the  mass  of  a  small  piece  of  a  body; 
then  N'dm  is  the  number  of  particles  in  dm.  If  x  is  the 
abscissa  of  dm,  then  x  X  N'dm  is  the  sum  of  the  abscissas 
of  all  the  particles  in  dm,  and  the  integral  J  Nx  •  dm  or  Nj  x  •  dm 
(extended  so  as  to  include  the  entire  body  under  discussion)  is 
the  sum  of  the  abscissas  of  all  the  particles  of  the  body.  Further- 
more NM  is  the  total  number  of  particles  in  the  body,  M  being 
the  mass  of  the  body.  Therefore,  dividing  Nfx-dm  by  NM 
we  have  the  sum  of  the  abscissas  of  all  the  particles  divided  by  the 
number  of  particles,  which  is  the  average  abscissa  of  the  entire  body. 
Representing  this  average  abscissa  by  X  we  have: 

NjX'dm      fx-dm 
^  ^      NM      ^  ~~M~~ 


MISCELLANEOUS  APPLICATIONS. 


129 


Therefore,  comparing  this  with  equation  (5)  of  Art.  74  we  see 

that  the  abscissa  of  the  center  of  gravity  of  a  body  is  the  average 

abscissa  of  all  the  particles  of  the  body,  and  similar  statements  can 

be  made  with  reference  to    Y  and   Z   as  given  by  equations  (6) 

and  (7)  of  Art.  74.     From  the  point  of  view  of  averages  the 

center  of  gravity  of  a  body  is  usually  called  the  center  of  mass 

of  the  body. 

PROBLEMS. 

1.  Four  downward  forces  of  100  pounds,  125  pounds,  200 
pounds  and  50  pounds  act  upon  a  bar  at  distances  of  10  inches, 
16  inches,  18  inches  and  24  inches  from  the  end  of  the  bar, 
respectively.  Find  the  value  of  the  resultant  of  the  four  forces 
and  the  distance  from  the  end  of  the  bar  to  the  point  of  applica- 
tion of  the  resultant.     Ans.  475  pounds,  16.4  inches. 

2.  Find  the  center  of  gravity  of  a  straight  bar  15  inches  long 
and  of  uniform  sectional  area  on  the  assumption  that  the  density 
of  the  material  is  given  by  the  equation  D  =  kx^;  where  D  is 
the  density  of  the  material  at  a  point,  x  is  the  distance  of  the 
point  from  the  end  A  of  the  rod,  and  fc  is  a  constant.  Ans.  12 
inches  from  end  A. 


B-\ i ^-. 


O 


1f-axi8 


circulitr  rod 


\ 


Fig.  pSa. 


Fig.  p36. 


3.  Find  the  location  of  the  center  of  gravity  of  a  rod  bent 
into  the  arc  of  a  circle  as  shown  in  Fig.  p3a.     Ans.  8.27  inches 
from  the  center  of  the  circle  in  the  line  AB, 
10 


130 


CALCULUS. 


Note,    Figure  p36  shows  how  this  problem  may  be  formulated.    The  length 

dx 

of  the  portion  ds  of  the  rod  is where  6  is  the  angle  between  ds  and 

^  cos »  * 

the  X-axis  and  it  is  equal  to  the  complement  of  the  angle  between  r  and  the 
X-axis.  That  is  cos  ^  =  -.  The  total  added  mass  corresponding  to  dx  is 
the  mass  of  the  two  short  portions  ds.    Therefore 


where  h  is  the  mass  of  the  rod  per  unit  length.    With  this  start  it  is  easy  to 
carry  the  problem  through  to  a  conclusion. 

4.  Find  the  distance  from  the  16  inch  base  to  the  center  of 
gravity  of  the  entire  wedge  (30  inches  long)  which  is  shown  in 
Fig.  p4.     Ans.  10  inches. 

5.  Find  the  distance  from  the  16  inch  base  to  the  center  of 
gravity  of  the  frustum  of  a  wedge  (12  inches  long)  which  is  shown 
in  Fig.  p4.    Ans.  5.5  inches. 


6.  Find  the  distance  from  the  apex  to*  the  center  of  gravity 
of  the  entire  cone  (10  inches  long)  shown  in  Fig.  pQ.  Ans.  7.5 
inches. 

7.  Find  the  distance  from  the  apex  to  the  center  of  gravity 
of  tlie  frustum  of  a  cone  (5  inches  long)  which  is  shown  in  Fig.  p6. 
Ans.  8.04  inches. 

8.  Find  the  position  of  the  center  of  gravity  of  a  solid  hemi- 


^^- 


MISCELLANEOUS  APPLICATIONS. 


131 


sphere  whose  radius   is    16    inches.     Ans.   6  inches  from  the 
center  of  the  sphere  on  the  axis  of  symmetry. 

Note.  This  problem  can  be  formulated  in  a  manner  very  similar  to  the 
formulation  of  the  problem  of  the  cone  which  is  discussed  in  Art.  76;  and  Fig. 
p8  will  suggest  the  method  of  formulation. 

9.  Figure  p9  represents  a  segment  of  a  solid  sphere.  Locate 
its  center  of  gravity.  Ans.  14.32  inches  from  the  center  of 
the  sphere  on  the  axis  of  symmetry  of  the  segment. 


y'^^^^y^segment  of  sphere 


(r-x) 


x-axis 


\ 

T 
■*—8lab  of  I 

thickness  dx     / 
/ 
/ 
/ 


Fig.  p8. 


I 


11 


Fig.  p9. 


10.  Locate  the  center  of  gravity  of  a  thin  hemispherical  shell 
whose  radius  is  16  inches.  Ans.  8  inches  from  the  center  of 
the  sphere  on  the  axis  of  symmetry  of  the  shell. 

iVote.  Let  Fig.  p36  represent  a  thin  spherical  shell.  Then  2-Ky.ds  is  the 
area  of  the  portion  of  the  shell  corresponding  to  dx.  The  mass  dm  of  this 
portion  of  the  shell  is  2-Ky  .ds  X  a,  where  a  is  the  mass  of  the  shell  in  pounds 
per  square  foot.  With  this  start  it  is  easy  to  carry  the  problem  through  to 
a  conclusion. 

1 1 .  Locate  the  center  of  gravity  of  the  solid  paraboloid  which 
is  shown  in  Fig.  pll.  Ans.  8  inches  from  the  vertex  of  the 
paraboloid  on  the  axis  of  the  paraboloid. 

12.  Locate  the  center  of  gravity  of  the  frustum  of  a  solid 


132 


CALCULUS. 


paraboloid  which  is  shown  in  Fig.  pl2.     Ans.  9.91  inches  from 
the  vertex  of  the  paraboloid  on  the  axis. 

13.  Locate  the  center  of  gravity  of  the  thin  paraboloidal  shell 


/ 

L7inch±8_ 

\ 


5  inches 


Fig.  pU. 


Fig.  pl2. 


which  is  shown  in  Fig.  pll.     Ans.  6.73  inches  from  the  vertex 
of  the  paraboloid  on  the  axis. 

Note.  In  the  solution  of  this  problem  it  is  necessary  to  integrate  an  ex- 
pression of  the  form  J  a:  Va  +  6aJ  •  dx.  This  expression  can  be  put  into  a 
form  whose  integral  is  easily  recognized  by  substituting       * 


z^  :=  a  -\-  bx 


80  that 


X  = 


dx  = 


z"  -  c 
h 

2zdz 


14.  Figure  pl4  represents  a  flat  metal  plate  cut  with  a  parabolic 
edge.  Locate  the  center  of  gravity  of  the  whole  plate.  Ans. 
5.4  inches  from  the  vertex  of  the  parabola  on  the  axis. 

15.  Locate  the  center  of  gravity  of  the  portion  AB  of  the 
plate  shown  in  Fig.  pl4.    Ans.  X  =  6.66  inches,    7  =  0. 

16.  Locate  the  center  of  gravity  of  the  portion  A  of  the  plate 
Bhown  in  Fig.  pl4.    Ans.  X  —  6.66  inches,    Y  =  3.42  inches. 

Note.  Consider  a  small  element  of  the  plate  as  represented  by  the  shaded 
square  in  Fig.  pl6.  The  area  of  this  element  is  dx.dy  and  its  mass  is  dm  = 
k'dx'dy,   where    &    is  a  confitant.     Therefore  the  integrals     fx-dm    and 


■au.^: 


MISCELLANEOUS  APPLICATIONS. 


133 


J  y-dm  become  kjjx'dx-dy  and  kjfy-dxdy,  respectively.  The  double 
integral  signs  indicate  that  two  integrations  are  necessary  in  each  case,  one 
with  respect  to  x  and  the  other  with  respect  to  y.  The  limits  of  these  integra- 
tions are  most  easily  assigned  when  the  integration  with  respect  to  y  is  made 


first.  In  this  case  the  integration  with  respect  to  y  is  between  the  Umits 
y  =  0  and  y  =  Vax  (where  y^  =  ax  is  the  equation  of  the  parabola).  In 
this  integration  x  and  dx  are  treated  as  constants.  Then  the  integration 
with  respect  to  a;  is  made  between  the  limits  x  =  4  inches  and  x  =  9  inches. 

17.  Locate  the  center  of  gravity  of  a  fiat  plate  which  is  in  the 
shape  of  a  sector  of  a  circle  as  shown  in  Fig.  pl7.  Ans.  In  the 
line   AB  at  a  distance  6.36  inches  from  A. 


y-axi8 


hemiapfiere 
thin  slab 


x^axia 


Fig.  pl7. 


Fig.  pl8. 


134  CALCULUS. 

18.  Suppose  that  the  density  of  the  material  of  which  a  sphere 
is  made  is  kr  where  A;  is  a  constant  and  r  is  any  distance  from 
the  center  of  the  sphere.  Locate  the  center  of  gravity  of  half 
of  the  sphere,  its  radius  being  2  feet.     Ans.  X  =  0.8  foot. 

Note.  Consider  a  ring-shaped  element  of  material  lying  in  the  thin  slab  in 
Fig.  pl8;  the  radius  of  the  ring  being  y  and  the  thickness  and  breadth  of  the 
ring  being  dx  and  dy  respectively.  The  volume  of  this  ring  is  ^irydx-dy 
and  its  mass  is  equal  to  kr  X  2iry'dx'dy,  where  r  =  Vx*  +  y^  as  shown 
in  the  figure.  Therefore  the  pull  of  gravity  (parallel  to  the  y-axis,  say)  on  the 
ring-shaped  element  is  2irky  Vx^  +  y^  -dx-dy  and  the  torque  action  of  this  pul 
about  the  axis  0  (perpendicular  to  the  plane  of  the  paper)  is: 


dT  =  2Trkxy\x^  +  y^' dx-dy 

With  this  start  the  problem  can  be  formulated  without  difficulty.  The 
total  mass  of  the  hemisphere  can  be  found  by  integrating,  between  proper 
limits,  the  expression: 

dM  =  2irky^x^  -{-  y^  'dx-dy 

19.  What  is  the  average  distance  of  the  points  on  a  semi-circle 
from  the  diameter  which  bounds  the  semi-circle?  Ans.  0.64 
of  the  radius. 

Note.  Let  N  be  the  nimiber  of  points  per  unit  length  of  a  line.  Then 
N-ds  \a  the  number  of  points  in  the  line  element  ds,  and  iVTrr  is  the  number 
of  points  on  the  semi-circle. 

20.  What  is  the  average  distance  of  all  the  points  in  a  semi- 
circular area  from  the  diameter  which  bounds  the  semi-circle? 
Ans.  0.42  of  the  radius. 

Note.    Let   N  represent  the  number  of  points  in  one  unit  of  area.    Then 

Ttr^ 
N-dA   is  the  number  of  points  in  an  element  of  area  dA   and  A^*  -7^-  is  the 

number  of  points  in  the  semi-circle. 

CENTER  OF  PRESSURE. 
78.  Force  exerted  on  a  water  gate.     Figure  68  shows  a  water 
gate  covering  an  opening  in  a  dam.     The  short  arrows  fff  repre- 
sent the  forces  with  which  the  water  pushes  against  the  gate. 
These  forces  are  together  equivalent  to  the  single  force  F,  their 


.^ 


MISCELLANEOUS  APPLICATIONS. 


135 


water  level     rrrrr 


dam 


Fig.  68. 

resultant,  the  point  of  application  of  which  is  called  the  center 
of  pressure  on  the  gate.  The  force  F  is  equal  to  the  sum  of  all 
the  forces  ///,  and  the  torque  action  of  F 
about  an  arbitrarily  chosen  axis  0  (see 
Fig.  69)  is  equal  to  the  sum  of  the  torque 
actions  about   0   of  the  forces  ///. 

Consider  a  horizontal  strip  of  the  gate 
of  which  the  width  is  dx  as  shown  in  Fig. 
69,  and  of  which  the  length  is  w.  The  area 
of  this  strip  is  w-dx  and  the  force  dF  ex- 
erted on  the  strip  is:* 

dF  =  Dwx-dx  (1) 

Therefore 

I       X'dx  (2) 

x=a  Fig.  69. 

*  The  pressure  in  pounds  per  square  foot  at  a  place  x  feet  beneath  the 
surface  of  water  is  p  =  Dx,  where  D  is  the  density  of  water  in  pounds  per 
cubic  foot  (=  62K).  Furthermore,  the  force  exerted  on  the  strip  is  found 
by  multiplying  the  area  of  the  strip  in  square  feet  by  the  pressure  in  pounds 
per  square  foot. 


136  CALCULUS. 

which  gives 

F  =  iDw{¥  -  a^)  (3) 

The  torque  action  of  the  force  dF  about  the  axis  0  is  x-dF 
or  Dwx^'dXf  and  the  total  torque  action  about  0  of  all  the 
forces  ///  is 

'~\^'dx  (4) 


T  =  DwT 

*Jx=a , 


which  gives 

T  =  \DwQy'  -  a^)  (5) 

This  torque  action  must  be  equal  to   XF,    Therefore  using  the 
value  of  F  from  equation  (3)  we  have 

\Dw{h''  -  a2)X  =  \Dw{W  ~  a^)  (6) 

whence 

79.  Force  exerted  on  a  curved  surface  by  a  stationary  fluid 
under  pressure.  The  short  arrows  ///  in  Fig.  70  represent 
the  forces  with  which  the  water  pushes  on  the  curved  surface  of 
a  dam.  All  the  forces  ///  are  together  equivalent  to  a  single 
force  F  (see  Fig.  73)  which  is  called  their  resultant.  Let  Fx 
and  Fy  be  the  x  and  y  components  of  F  as  shown  in  Fig.  73. 
Then  Fx  is  the  sum  of  the  a:-components  of  the  forces  fff 
in  Fig.  70,  and  Fy  is  the  sum  of  the  ^/-components  of  the 
forces  ///.  Let  X  and  Y  be  the  coordinates  of  the  point  of 
appHcationof  the  resultant  force  F  (see  Fig.  73).  Then  we  have 
the  following  four  conditions  which  determine  Fxy  Fy,  X  and  Y: 

(a)    Fx  is  the  sum  of  the  aj-components  of  the  forces  fff. 

(h)     Fy  is  the  sum  of  the  ^/-components  of  the  forces  fff. 

(c)  The  torque  action  of  Fx  about  the  arbitrarily  chosen 
axis  0  is  FxY,  as  may  be  seen  from  Fig.  73,  ahd  this  torque 
action  is  equal  to  the  sum  of  the  torque  actions  about  0  of 
the  a;-components  of  the  forces  ///. 


MISCELLANEOUS  APPLICATIONS. 


137 


(d)  The  torque  action  of  Fy  about  0  is  FyX,  and  this 
torque  action  is  equal  to  sum  of  the  torque  actions  about  0  of 
the  ^/-components  of  the  forces  ///. 

In  order  to  formulate  these  four  conditions  it  is  necessary  to 
derive  expressions  for  the  x  and  y  components  of  the  force  dF 
which  is  exerted  by  a  fluid  on  an  element  of  a  curved  surface. 
Consider  the  surface  element  AB,  Fig.  71,  the  width  of  the 
element  being  ds  and  its  length  I  (perpendicular  to  the  plane 
of  the  paper).  The  area  of  AB  is  l.ds  square  feet,  and  the  force 
in  pounds  exerted  on  AB  is  dF  =  pl.dSf  where  p  is  the  pressure 


,umter  level 


Fig.  70. 


of  the  fluid  in  pounds  per  square  foot.     The  x  and  y  components 
of  dF  are,  respectively: 


and 


dF:,  =  sin0-dF 
dFy  =  cos  (f)'dF 


dx 


Therefore,  using  pl-ds  for  dF,  using  -p  for  cos  0,   and  using 

-r   for  sin  6,  we  have 
ds 


and 


dF.  ==pl-dy  (1) 

dFy  =  phdx  (2) 

To  determine  the  total  force  exerted  on  the  face  of  the  dam  in 


138 


CALCULUS. 


Fig.  70,  that  is,  to  determine  the  two  components  Fx  and  Fy 
of  the  total  force  and  the  coordinates  X  and  Y  of  its  point  of 
appHcation  as  shown  in  Fig.  73,  consider  the  element  of  area 
hds   as  shown  in  Fig.  72,  where   I  is  the  length  of  the  dam  per- 


pendicular to  the  plane  of  the  paper.  The  pressure  of  the  water 
at  this  element  is  Dy  pounds  per  square  foot  as  explained  in 
the  footnote  to  Art.  78.  Therefore,  using  Dy  for  p  in  equations 
(1)  and  (2),  we  have  for  the  x  and  y  components  of  the  force 
dF  in  Fig.  72: 

dFx  =  Dlydy  (3) 

and 

dFy  =  Dlydx  (4) 

From  equation  (3)  we  find  the  x-component  of  the  total  force 
exerted  on  the  dam  by  integrating  between  the  limits  y  =  Q 
to  y  =  hj  which  gives: 

F,  =  iDlh^  (5) 

The  equation  of  the  parabola  in  Fig.  72  is 

^         2 

y  =  p-^^ 

Therefore,  using  this  value  of  y  in  equation  (4)  and  integrating 


MISCELLANEOUS  APPLICATIONS.  139 

between  the  limits   x  =  0   to    x  =  h,    we  find  the  2/-component 
of  the  total  force  exerted  on  the  dam,  namely: 

Fy  =  \Dlhh  (6) 

To  find  the  distance  X  in  Fig.  73  it  is  necessary  to  find  the 
total  torque  action  about  0  of  all  the  forces  dFy  and  place  the 
result  equal  to  FyX.  The  torque  action  of  dFy  (the  i/-com- 
ponent  of  dF  in  Fig.  73)  about  0  is  a; .  dFyj  which  by  equation 
(4)  becomes  Dlxy-dx.     But 

so  that  the  torque  action  of  dFy  about  0  is  DlroOc^-dx,  and  the 
torque  action  of  all  the  forces   dFy   about   0   is 


Dlrh  I       x^-dx 

O^Jx=0 


which  is  equal  to  \I)lhh^.     Therefore  we  have 

XFy  =  \DlhW  (7) 

or,  using  the  value  of  Fy  from  (6),  we  have: 

X  =  16  (8) 

The  torque  action  of  dFx  (the  x-component  of  dF  in  Fig.  73) 
about  0  is  y-dFx,  which  by  equation  (3)  becomes  Dly^-dy; 
and  the  total  torque  action  of  all  the  forces  dFx  is 


Jf*V=n 


which  is  equal  to  iDl¥.    Therefore  we  have: 

YFx  =  iDW  (9) 

or,  using  the  value  of  Fx  from  (5),  we  have 

Y  =  ih  (10) 


140 


CALCULUS. 


y-axis 


PROBLEMS. 

1.  Find  the  total  force  F  acting  on  the  gate  in  Fig.  68  and 
find  the  distance  X,  when  a  =  4  feet,  6  =  8  feet  and  ly  =  4 
feet.    Ans.  6000  pounds,  6.22  feet. 

2.  A  dam  has  a  circular  hole  through 
it  6  feet  in  diameter.  The  center  of 
the  hole  is  7  feet  beneath  the  surface  of 
the  water.  The  hole  is  covered  with 
a  gate.  Find  the  total  force  with 
_  which  the  water   pushes  on   the   gate 

4---(vr/77r^  1         ^^^  fiiid  the  distance  from  the  surface 

I      V///////1    "V  r       I 

of  the  water  to  point  of  application 
of  this  force.  Ans.  12,370  pounds,  7.32 
feet. 

X-axis  Note.    Consider  the  element  of  area  which  is 

»  represented  by  the  small  shaded  square  m  Fig.  p2. 

^^'  ^  •  The  pressure  at  this  area  is    kx    where    A;    is  a 

constant.    Therefore  the  force  acting  on  the  element  of  area  is    kx-dx'dy, 

and  the  total  force  on  the  gate  is    jj  kx'dx'dy.    To  specify  the  limits  of  this 

integration  let  y  =  f(x)  be  the  equation  of  the  circle  in  Fig.  p2  (the  gate). 
Then  if  we  integrate  first  with  respect  to  y  the  limits  are  y  =  —  fix)  to 
y  =  +/(x).  We  may  then  integrate  with  respect  to  x  between  the  Umits 
X  =  h  —  r  to  x  =  6  +  r. 

The  torque  action  of  the  force  kx-dx'dy  about  the  axis  0  (perpendicular 
to  the  plane  of  the  paper)  is    kx^-dx-dy,    and  the  total  torque  action  of  the 

force  acting  on  the  entire  gate  is  given  by  the  integral  J  J  kx^'dx-dy.  The 
limits  are  the  same  as  stated  above. 

3.  Find  the  abscissa  X  and  the  ordinate  Y  of  the  center  of 
pressure  of  the  shaded  half  of  the  circular  water  gate  in  Fig.  p2, 
where  6  =  7  feet  and  r  =  6  feet.  Ans.  X  =  7.32  feet,  Y  = 
1.27  feet. 

MOMENT  OF  INERTIA. 

80.  Kinetic  energy  of  a  rotating  body.  Definition  of  moment 
of  inertia.     Consider  a  wheel  which  is  rotating    n    revolutions 


MISCELLANEOUS  APPLICATIONS.  141 

per  second.     Its  speed  in  radians  per  second  is 

CO  =  2irn  (1) 

because  there  are  2%  radians  in  one  revolution.  The  speed  of 
a  rotating  body  in  radians  per  second  is  called  the  angular 
velocity  or  the  spin  velocity  of  the  body. 

Consider  a  particle  of  the  wheel  at  a  distance  r  from  the  axis 
of  rotation.  As  the  wheel  rotates  this  particle  travels  in  a  circle 
of  which  the  circumference  is  27rr  and  it  travels  n  times  round 
this  circle  per  second  so  that  the  velocity    v    of  the  particle  is 

V  =  2Trnr  (2) 

or  using  w  for  27rn   according  to  equation  (1),  we  have: 

v  =  ojr  (3) 

If  the  spin  velocity  of  the  wheel  is  doubled  it  is  evident  from 
this  equation  that  the  velocity  of  every  particle  in  the  wheel 
will  be  doubled  so  that  the  kinetic  energy  of  every  particle  in  the 
wheel  will  be  quadrupled.  Therefore  the  total  kinetic  energy, 
W,  of  a  rotating  wheel  is  quadrupled  if  the  spin  velocity  co 
of  the  wheel  is  doubled;  that  is,  for  a  given  wheel,  W  is  pro- 
portional to  £0^,  and  for  the  given  wheel  there  is  a  definite 
constant  by  which   co^   may  be  multiplied  to  give   W.     That  is: 

W  =  iKo)''  (4) 

where  (^K)  is  the  proportionality  factor  for  the  given  wheel. 
The  constant  K  is  called  the  moment  of  inertia  of  the  wheel,*  and 
it  depends  upon  the  size,  shape,  and  mass  of  the  wheel. 

*  The  kinetic  energy  of  a  particle  is  equal  to  ^mv^,  where  m  is  the  mass 
of  the  particle  in  pounds,  v  is  its  velocity  in  feet  per  second,  and  kinetic  energy 
is  expressed  not  in  foot-pounds,  but  in  foot-poundals.  Throughout  this  dis- 
cussion of  moment  of  inertia  distance  is  expressed  in  feet,  velocity  in  feet  per 
second,  mass  in  pounds,  force  in  poundals,  torque  in  poundal-feet,  work  or 
energy  in  foot-poundals,  and  moment  of  inertia  in  pound-feet.^ 


142  CALCULUS. 

81.  General  integral  expression  for  moment  of  inertia.  Imag- 
ine a  small  particle  of  mass  dm  to  be  added  to  the  spinning 
body  at  a  distance  r  from  the  axis  of  spin.  Then  the  velocity 
V  of  the  added  particle  is  cor,  and  the  kinetic  energy  of  the  added 
particle  is  ^dm  X  coV  which  is,  of  course,  the  infinitesimal 
increment  of  the  kinetic  energy  of  the  spinning  body  due  to  the 
added  particle.     Therefore 

dW  =  ioy^rHm 
and  by  integration*  we  have 

W  =  Wfr^-dm  (5) 

Comparing  this  equation  with  equation  (4)  it  is  evident  that 

K^fr^'dm  (6) 

82.  Average  value  of  the  square  of  the  distances  of  all  the 
particles  of  a  body  from  an  axis.     Definition  of  radius  of  gyration. 

Using  the  ideas  of  Art.  77,  N-dm  is  the  number  of  particles  in  a 
small  piece  dm  of  a  body.  If  r  is  the  distance  of  the  small 
piece  dm  from  a  chosen  axis,  then  Nr^-dm  is  the  sum  of  the 
squares  of  the  distances  of  all  the  particles  in  dm  from  the  axis, 
and  the  integral  jNr^.dm  or  Nfr^-dm  extended  so  as  to 
include  an  entire  body  is  the  sum  of  the  squares  of  the  distances 
of  all  the  particles  of  the  body  from  the  axis.  But  NM  is 
the  number  of  particles  in  the  body.     Therefore 

N  f  r^ '  dm       fr^  •  dm 

— ^ =  ^ (I) 

is  the  average  value  of  the  squares  of  the  distances  of  all  the 
particles  of  a  body  from  the  axis.     This  average  may  be  repre- 

*  This  integration  can  only  be  indicated.  Before  the  actual  value  of  the 
integral  can  be  found  r  and  dm  must  be  expressed  in  terms  of  one  or  more 
independent  variables,  and  the  limits  of  the  integration  must  be  such  as  to 
include  the  entire  spinning  body. 


MISCELLANEOUS  APPLICATIONS. 


143 


sented  by   p^  so  that 


f  r^  •  dm 


M 


or 


p2M  =  fr^-dm 


(2) 


(3) 


but  J  7-2  •  dm  is  the  moment  of  inertia  K  of  the  body  with  respect 
to  the  chosen  axis,  according  to  equation  (6)  of  Art.  81.  There- 
fore we  have: 

K  =  pm  (4) 

The  distance  p,  which  is  the  square-root-of-thc-aVerage-value- 
of-the-squares-of-the-distances-of-all-the-particles-of-a-body-from- 
a-chosen-axis,  is  called  the  radius  of  gyration  of  the  body  with 
respect  to  the  chosen  axis. 

83.  Moment  of  inertia  of  a  circular  saw  about  its  axis  of 
spin.  The  moment  of  inertia  of  a  circular  saw  could  be  easily 
derived  from  equation  (6)  of  Art.  81,  but  it  is  instructive  to  give 
the  complete  argument  again  as  follows:  Let  W  be  the  kinetic 
energy  of  a  circular  disk  r  feet  in  radius  rotating  at  a  constant 
spin  velocity  of   co   radians  per  second,  as  shown  in  Fig.  74,  and 


n— -.^- 


added      , 
material 


Fig.  74. 

let  it  be  required  to  find  the  infinitesimal  increment  of  W  due 
to  an  arbitrary  infinitesimal  increment  of  r.  Let  t  be  the  thick- 
ness of  the  disk  and  let  D  pounds  per  cubic  foot  be  the  density 
of  the  material  of  which  the  disk  is  made.     Imagine  the  radius 


144  CALCULUS. 

of  the  spinning  disk  to  be  increased  by  the  addition  of  material 
as  indicated  in  Fig.  74.  The  volume  of  the  added  material  is 
2irr  X  t  X  dr,  the  mass  of  the  added  material  is  2-Ktr'dr  X  D 
pounds,  the  velocity  of  the  added  material  is  or,  the  kinetic 
energy  of  the  added  material  is  equal  to  one  half  the  product  of 
its  mass  times  the  square  of  its  velocity,  and  the  kinetic  energy 
of  the  added  material  is  the  desired  infinitesimal  increment  dW. 
Therefore : 

dW  =  rDUa'^r^'dr  (1) 

W  and  r  being  the  only  variables.  Therefore,  by  integration 
we  get: 

W  =  JttD^coV^  +  a  constant  (2) 

But  W  must  evidently  be  zero  when  r  is  zero.  Therefore  the 
constant  of  integration  must  be  equal  to  zero,  so  that  equation 
(2)  becomes 

W  =  iirDtJ'r'  (3) 

Now  irrHD  is  equal  to  the  mass  m  of  the  disk  in  pounds  so  that 
equation  (3)  becomes: 

W  =  iw^mr^  (4) 

But  according  to  Art.  80  the  kinetic  energy  of  any  rotating  body 
can  be  expressed  as  ^Ku^,  where  K  is  the  moment  of  inertia 
of  the  body.    Therefore  we  have 

^v  W  =  i«2mr2  =  ^Kc^  (5) 

from  which  we  have: 

K  =  imr^  (6) 

That  is,  the  moment  of  inertia  of  a  circular  saw  about  its  axis 
of  spin  is  equal  to  one  half  the  mass  of  the  saw  in  pounds  multi- 
plied by  the  square  of  the  radius  of  the  saw. 

Remark.  The  thickness  i  in  the  above  discussion  may  be 
anything  whatever.  Therefore  equation  (6)  expresses  the  mo- 
ment of  inertia  of  a  circular  cylinder  of  any  length  rotating  about 
its  axis  of  figure. 


MISCELLANEOUS  APPLICATIONS. 


145 


84.  Moment  of  inertia  of  a  rectangular  bar.  To  determine 
the  moment  of  inertia  of  the  rectangular  bar  shown  in  Fig.  75 
the  general  equation  (6)  of 
Art.  81  will  be  used.  Figure 
76  represents  a  top  view  of 
the  bar,  0  being  the  axis  of 
rotation.  Consider  the  ele- 
ment of  the  barwhich  is  rep- 
resented by  the  small  black 
square  in  Fig.  76.  This  ele- 
ment is  understood  to  ex- 
tend entirely  through  the 
bar  parallel  to  the  axis  0 
paper  in  Fig.  76). 


Fig.  75. 


(at  right  angles  to  the  plane  of  the 
Therefore  the  volume  of  the  element  is 
t'dx'dy  and  its  mass  is  tD-dx'  dy.  The  distance  of  the  element 
from  the  axis  is   r  =  Vx^  +  y^,   so  that 

r^  =  x^  +  y^ 

Therefore  the  general  equation  (6)  of  Art.  81  becomes: 

K  =  tnffix^  +  2/2)  'dx'dy  (1) 

in  which  the  double  integral  sign  is  used  because  two  integrations, 
one  with  respect  to  x  and  one  with  respect  to  y  are  necessary, 
as  explained  in  the  note  to  problem  16  on  page  132. 

If  we  integrate  with  respect  to  y  (treating  x   and  dx  as  con- 
stants) between  the  limits 
L 


y  = 


to 


t/  =  +  -  we 


Fig.  76. 


get  the  moment  of  in- 
ertia with  respect  to  the 
axis  0  of  the  thin  slab  be- 
tween the  dotted  lines  in 
Fig.  76.  We  may  then  inte- 
l     .  .   I 


grate  with  respect  to  x  between  the  limits  x  =  —-^    to    x 

11 


=  +  2 


146  CALCULUS. 

and  we  get  the  desired  expression  for  K,   namely: 

K  =  ^\V  +  ^)  (2) 

But  Dtwl  is  the  mass  m  of  the  bar  in  pounds,  so  that  equation 
(2)  becomes: 

X=^(P +  «,»)«  (3) 

Remark.  The  same  final  result  is  obtained  if  we  integrate 
first  with  respect  to  x  (treating  y  and  dy  as  constants),  and 
then  with  respect  to  y.  What  is  the  meaning  of  the  result  of 
the  first  integration  in  this  case? 

85.  Torque  required  to  increase  the  spin- velocity  of  a  body. 

The  velocity  of  a  particle  in  a  rotating  wheel  is  t;  =  rw,  according 
to  equation  (3)  of  Art.  80.  Therefore  when  «  increases,  v 
increases  r  times  as  fast  as   w.     That  is: 

dv        du)  ,^v 

This  -r.  is  the  acceleration  of  the  particle  in  the  direction  at 
right  angles  to  r*,   and 

dv  J     /        d(a    J   \ 

is  the  sidewise  force  (at  right  angles  to  r)  which  must  act  on  the 
particle  to  produce  the  sidewise  acceleration.  The  torque  action 
of  this  sidewise  force  about  the  axis  of  rotation  is: 

dT  =  r^'dmXr 
dt 

or 

dT  =  ~  -r'-dm  (2) 

*  We  are  not  here  concerned  with  the  radial  acceleration  of  the  particle 
which  is  discussed  in  Art.  50. 


MISCELLANEOUS  APPLICATIONS.  147 

so  that  the  total  torque  action  required  to  increase  the  spin 
velocity  of  the  body  at  the  rate   -t:    is : 

^''^r'^-dm  (3) 


T  =  —  C 
dtj 


The  integral  is  of  course  understood  to  be  extended  over  the 
whole  body  and  it  is  equal  to  the  moment  of  inertia  of  the  body 
according  to  equation  (6)  of  Art.  81.  Therefore  equation  (3) 
may  be  written 

r  =  Kf  (4) 

That  is,  the  spin  acceleration   -^    of  a  body  multiplied  by  the 

moment  of  inertia  of  the  body  is  equal  to  the  torque  which  must 
be  exerted  on  the  body  to  produce  the  spin  acceleration. 

86.  Moments  of  inertia  about  par-  ^dsf 

allel  axes.     Let   K    be  the  moment  ^y"^ 

of  inertia  of  a  body   about  an  axis  ^^^     A 

(perpendicular   to   the   plane   of  the  ^^  / 1 

paper  in  Fig.  77)  which  passes  through      *i-.^^2 ^-. 

the  center  of  gravity    0   of  the  body,  pjg  77^ 

and  let  K'  be  the  moment  of  inertia 

of  the  body  about  an  axis  P    (perpendicular  to  the  plane  of  the 

paper)  which  is  at  a  distance  a  from   0.     Then 

K'  =  Z  +  aW  (1) 

where  M  is  the  total  mass  of  the  body  in  grams  or  pounds  as  the 
case  may  be. 

Proof.     From  the  triangle  in  Fig.  77  we  have: 

gf2  =  (j2  _j_  j,2  _j_  2ar  cos  B 
or 

52  =  (j2  4.  y.2  _|_  2ax  (2) 

where  x  is  the  abscissa  of  the  element  of  material  dm  referred 


148  CALCULUS. 

to  the  center  of  gravity  0  as  an  origin.  Now  according  to 
equation  (6)  of  Art.  81  we  have 

K'  =  fq'-dm  (3) 

Therefore,  using  the  value  of  q^  from  equation  (2),  we  have: 

K'  =  fa^-dm  +  fr^-dm  +  f2ax-dm 

K'  =  a'fdm  +  Jr^-dm  +  2afx'dm  (4) 

But  J  dm    =  the  total  mass   M   of  the  body,   jr^-dm  is  the 

moment  of  inertia  K  referred  to  the  axis  0,  and  jx-dm  is 
zero  according  to  equation  (5)  of  Art.  74,  because  the  center  of 
gravity  0  in  Fig.  77  is  taken  as  the  origin.  Therefore  equation 
(4)  becomes: 

K'  =  am  +  K  (1) 

87.  Moment  of  section  of  a  beam.     Consider  equation  (2)  of 
Art.  72,  namely: 

dT  =  ^^x^'dx  (1) 

The  product  2b -dx  is  the  area  of  the  shaded  strips  in  Fig.  59,  and 
X  is  the  distance  of  this  area  from  the  axis  00,  Let  us  use  dA 
for  2b 'dx,  then  equation  (1)  becomes: 

dT  =  ^x^'dA  (2) 

80  that 

^  "*  :^'dA  (3) 


-f/' 


The  integral  fx*  •  dA  is  called  the  moment  of  section  of  the  beam; 
and  according  to  equation  (3)  the  torque    T    which  bends  any 

beam  is  equal  to   -h-  »   where   E    is  the  stretch  modulus  of  the 
K 

material  of  the  beam,  R  is  the  radius  of  curvature  of  the  median 
line  of  the  beam,  and  S[  =fx*  •  dA)  is  the  moment  of  section  of 
the  beam. 


MISCELLANEOUS  APPLICATIONS. 


149 


The  integral  fx^-dA  is  similar  in  form  to  the  integral  fr^-dm 
in  equation  (6)  of  Art.  81,  and  because  of  this  similarity  engineers 
call  J x^ '  dA  the  "moment  of  inertia  "  of  the  section  of  the  beam. 
The  term  wmnent  of  section  is  however  the  correct  term. 


PROBLEMS. 

1.  Find  the  moment  of  inertia  of  a  circular  saw  six  feet  in 
diameter  and  of  which  the  mass  is  125  pounds.     Ans.  562.5  Ib.-ft.^ 

2.  Find  the  amount  of  energy  in  foot-pounds  stored  in  the 
saw  of  problem  1  at  a  speed  of  600  revolutions  per  minute.  Ans. 
34482  foot-pounds. 

Note.  In  the  formula  W  =  }4Ku^,  K  is  expressed  in  pound-feet-squared, 
w  is  expressed  in  radians  per  second,  and  W  is  expressed  in  foot-poundals 
(not  in  foot-pounds).     There  are  32.2  foot-poundals  in  one  foot-pound. 

3.  Find  the  moment  of  inertia  of  a  cylinder  2  feet  in  diameter 
and  3  feet  long  with  respect  to  its  axis,  the  density  of  the  material 
being  420  pounds  per  cubic  foot.     Ans.  1980  Ib.-ft.^ 

4.  Find  the  moment  of  inertia  of  a  hollow-cylinder,  the  external 
dimensions  being  the  same  as  in  problem  3,  inside  diameter  being 
1  foot,  the  axis  of  revolution  being  the  axis  of  the  cylinder,  and 
the  density  of  the  material  being  420  pounds  per  cubic  foot. 
Ans.  1114  lb.-ft.2 

5.  Find  the  moment  of  inertia  of 
the  rectangular  bar  in  Figs.  75  and 
76.  The  bar  is  5  feet  long,  1  foot 
wide  and  0.5  foot  thick  and  it  has 
a  mass  of  1000  pounds.  Ans. 
2167  lb.-ft.2 

6.  Find  the  moment  of  inertia  of 
the  bar  in  problem  5  when  the  axis 
of  rotation  coincides  with  the  edge 
t    in  Fig.  75.     Ans.     8667  Ib.-ft.^ 

7.  Find  the  moment  of  inertia 
of  a  very  thin  circular  disk  when 


If-axis 
,^axi8pf^  revolution 


x-axi8 


Fig.  p7. 


150 


CALCULUS. 


the  axis  of  rotation  is  a  diameter  of  the  disk;  the  mass  of  the  disk 
being  120  pounds  and  its  diameter  being  6  feet.    Ans.  270  Ib.-ft.^ 

Note.  Let  the  circle  in  Fig.  p7  represent  the  disk.  Take  the  narrow 
vertical  black  strip  as  dm.  Then  dm  =  k  X2y-dx  where  A;  is  pounds  per 
square  foot  of  disk  area. 

8.  Find  the  moment  of  inertia  of  the  thin  circular  disk  with 
respect  to  the  axis    0    as  shown  in  Fig.  p8,  the  density  of  the 


front  view 


Fig.  p8. 


/  7?2         \ 
material  being  D.    Ans.  K  =  tR^D  l-j-^x^j-dx. 

Note.    The  mass  of  the  narrow  strip  in  Fig.  p8  is 


2D  a//?  _  y^'dx'dy  =  dm 
and  the  distance  of  the  strip  from  the  axis  is 


Va:»  +  2/2  =  r 
Of  course  x  and  dx  are  constants  in  this  problem. 


30  inches 

Fig.  p9. 


hoop  3  feet 
in  diameter 


Fig.  plO. 


MISCELLANEOUS  APPLICATIONS. 


151 


9.  Find  the  moment  of  inertia  of  the  soUd  cylinder  shown  in 
Fig.  p9;  the  density  of  the  material  being  0.28  pound  per  cubic 
inch.     Ans.  1833  Ib.-ft.^ 

Note.    The  method  of  formulating  this  problem  is  suggested  by  problem  8. 

10.  A  slender  circular  ring  or  hoop  has  a  mass  of  10  pounds 
and  it  is  3  feet  in  diameter,  (a)  Find  its  moment  of  inertia 
about  the  axis  C  (perpendicular  to  the  plane  of  the  paper  in  Fig. 
plO),  and  (6)  find  its  moment  of  inertia  about  the  axis  ab. 
Ans.  (a)  22.5  Ib.-ft.^  (6)  11.25 
lb.-ft.2 

11.  Find  the  moment  of  in- 
ertia of  a  solid  steel  sphere  10 
inches  in  diameter,  the  axis  of 
revolution  being  a  diameter  of 
the  sphere.  The  density  of  steel 
is  0.28  pound  per  cubic  inch. 
Ans.  22.4  Ib.-ft.^ 


cuds  of  revolution 
solid  sphere 


cylindrical 
shell 


Note.     Let  the   circle  m  Fig.  pll 

represent  the  sphere.     Take  for  dm 

a  thin  cylindrical  shell  of  radius  r  and 
thickness  dr. 


Fig.  pll. 


12.  Find  the  moment  of  inertia  of  a  steel  governor  ball,  4 
inches  in  diameter,  with  respect  to  the  axis  of  revolution  of  the 
governor,  the  center  of  the  ball  being  6  inches  from  the  axis. 
Ans.  2.45  Ib.-ft.^ 


TABLE  OF  MOMENTS  OF  INERTIA. 

Axis  through  center  of  mass  in  each  case, 
m  =  mass  of  body  in  grams  or  pounds. 
K  =  moment  of  inertia. 


I    and  having  uniform  section, 


1.  Thin  straight  bar  of  length 
axis  at  right  angles  to  bar.     K  ■■ 

2.  Rectangular  parallelopiped,  axis  parallel  to  one  edge, 
and  b  being  lengths  of  other  edges.    K  =  j^ia^  +  b'^)m 


a 


152  CALCULUS. 

3.  Cylinder  or  disk  of  radius    r,    referred  to  axis  of  cylinder. 
K  =  ir^m. 
Referred  to  axis  at  right  angles  to  axis  of  cylinder 


--(5+0 


m 


where  I  is  the  length  of  the  cylinder. 

4.  Hollow  cylinder  oirsidu  R  and  r,  referred  to  axis  of  cylinder. 
K  =  ^{R^  +  r^)m. 

5.  Sphere  of  radius  r  referred  to  a  diameter.     K  —  \r^m. 

Note.    If  the  axis  of  rotation  does  not  pass  through  the  center  of  mass,  use 
equation  (1)  of  Art.  86. 


CHAPTER  VI. 

EXPANSIONS  IN  SERIES.    USE  OF  COMPLEX  QUANTITY. 

88.  Maclaurin's  theorem. — Let  y  be  any  function  whatever 
of  X  which  is  finite  and  continuous  and  of  which  all  of  the 
derivatives  with  respect  to  x  are  finite  and  continuous.     Then: 

y  =  A4-Bx+^Cx2  +  -ilDx«+|^Ex^+...  (1) 

where  A,  B,  C,  D,   etc.,  are  constants  as  follows: 


A 

is  the  value  of    y 

when  X 

=  0 

B 

is  the  value  of   y^ 
dx 

when  X 

=  0 

C 

is  the  value  of  -r^ 

when  X 

=  0 

D 

is  the  value  of    ,  , 

when  X 

=  0 

E 

is  the  value  of  ^ 

when  X 

=  0 

etc., 

etc. 

Equation  (1)  expresses  what  is  known  as  Maclaurin^s  theorem. 

Proof. — Let  the  curve  cc  in  Fig.  78  represent  the  given 
function.  Then  the  value  of  y  which  is  to  be  expressed  by 
equation  (1)  is  the  ordinate  of  the  point  p.  To  establish  equa- 
tion (1)  we  will  make  a  series  of  approximations  and  consider 
the  limit  towards  which  this  series  of  approximations  trends, 
as  follows: 

First  approximation. — To  get  a  first  approximation  let  us 

dv  dv 

assume  that    -^    is  equal  to  the  constant    B    (the  value  of    -^ 

153 


154 


CALCULUS. 


when  X  =  0)   everywhere  between  the  points  p  and  q  in  Fig. 
78.     That  is  by  assumption  we  have: 


dx 


^  B 


(2) 


Integrating  this  differential  equation  we  have: 

y  =  Bx  -\-  a.  constant 
but  y  =  A  when  x  =  0,  so  that  the  constant  of  integration  is 


Fig.  78. 


evidently  equal  to  A.  Therefore  as  a  first  approximation  we 
have: 

y  =  A  +  Bx  (3)* 

dv 
It  is  interesting  to  note  that  to  assume    -^  =  B    everywhere  is 

the  same  thing  as  to  take  the  inclined  dotted  line  in  Fig.  78  as 
an  approximation  to  the  curve  cc;  and  the  ordinate  of  this 
inclined  straight  line  is  A  +  Bx. 


*  Figure  79  shows  a  curve    pq    for  which 


dy 

dx 


B    everywhere,  but  the 


ordinate  of  p  is  not  equal  to  A  -{■  Bx  because  of  the  discontinuity  or  jump 
at  d.  The  function  y  and  all  of  its  derivatives  must  be  finite  and  continuous 
everywhere  between  p  and  q,  as  stated  at  the  beginning.  Any  discontinuity 
outside  of  the  region  between  p  and  q  does  not  vitiate  equation  (1)  for  the 
region  between  p  and  q. 


EXPANSIONS  IN  SERIES.  155 

Second  approximation. — To  get  a  second  approximation  let 
us  assume  that    -j-^    is  equal  to  the  constant    C    (the  value  of 

-T^    when    a;  =  0)    everywhere  between    p    and    q    in  Fig.  78. 
That  is,  by  assumption,  we  have: 

d^y 


Integrating  once  we  have: 


d.^  =  ^  w 


^  =  Cx  +  a  constant 
dx 

dii 
But   -f  —  B   when   a;  =  0,   so  that  the  constant  of  integration 

is  equal  to   5,   giving: 

%  =  B  +  Cx  (5) 

Integrating  again  we  have : 

y  =  Bx  +  -x  Cx^  +  a  constant 

But    y  =  A    when    x  =  0,    so  that  the  constant  of  integration 
is  equal  to  A,   giving  as  a  second  approximation: 

y  =  A  +  Bx  +  ^Cx2  (6) 

Third  approximation. — To  get  a  third  approximation  let  us 

assume  that   -r^   is  equal  to  the  constant   D    (the  value  of  -~ 

when  X  =  0)  everywhere  between  p  and  q  in  Fig.  78.     That  is, 
by  assumption,  we  have: 

By  three  successive  integrations  (the  constant  of  each  Integra- 


156  CALCULUS. 

tion  being  determined  as  above)  we  get,  as  a  third  approximation: 

y  =  A  +  Bx4-|cx2  +  iDx3  (8) 

nth  approximation. — To   get   an   nth   approximation   let  us 

assume  that    7-^   is  equal  to  the  constant   N   (the  value  of  t^ 

when  a;  =  0)  everywhere  between  p  and  q  in  Fig.  78.  That  is, 
by  assumption,  we  have: 

g  =  ^  (9) 

By  n  successive  integrations  this  differential  equation  gives, 
as  the  nth  approximation: 

y=  A  +  Bx  +^Cx2  +  ||Dx3  +  . . .  +  j^Nx"         (10) 

To  show  that  the  nth  approximation  approaches  the  true  value  of 
y  as  a  limit  as  n  approaches  infinity.  In  the  first  place  it  is 
evident  that  equation  (10)  gives  equation  (1)  when  n  is  in- 
creased more  and  more,  but  it  remains  to  be  shown,  that  y  in 
equation  (10)  approaches  the  correct  value  of  ?/  as  a  limit  as  n 
is  increased. 

d^ti 
The  nth  derivative    -t\    is  assumed  to  be  everywhere  finite 

and  it  must  have  therefore  a  definite  largest  value  V  and  a 
definite  smallest  value  v  between  p  and  q.  If  the  largest  value 
V  is  used  instead  of  N  in  equation  (10)  we  get  too  large  a  value, 
a,  for  y.  If  the  smallest  value  v  is  used  instead  of  N  in  equa- 
tion (10)  we  get  too  small  a  value,   b,   for  y.*    That  is: 

a  =  A  +  Bx+~Cx^+"'n'Vx''  (11) 

*  This  statement  happens  to  be  plausible  and  therefore  the  reader  is  apt 
to  accept  it  as  true  without  actually  perceiving  its  truth,  which  is  indeed 
evident  if  one  takes  the  trouble  to  think  about  it. 


EXPANSIONS  IN  SERIES.  157 

and 

h  =  A  -i-  Rv  4-     ^^    ,        

2         '  1^ 


A  +  Bx+lcx"  +  . . .  ^.  t;x"  (12) 


and  the  true  value  of  y  lies  between  a  and  b.     But,  subtracting 
equation  (12)  from  equation  (11)  member  by  member  we  get: 


and  this  difference  approaches  zero  as  n  approaches  infinity, 
because  when  n  is  increased  by  1  the  numerator  is  multiplied 
by  the  finite  quantity  x,  whereas  the  denominator  is  multipUed 
by  the  quantity  n  +  1,  which  becomes  as  large  as  you  please. 
It  is  evident  therefore  that  the  value  of  a  as  given  by  equation 
(11)  becomes  more  and  more  nearly  equal  to  the  value  of  6 
as  given  by  equation  (12).  But  the  true  value  of  y  Hes  between 
a  and  h.  Therefore  equations  (11)  and  (12)  both  approach  the 
true  value  oi"  y  as  a  limit  as  n  is  increased;  and  equations  (11) 
and  (12)  reduce  to  equation  (1)  when  n  is  indefinitely  great. 

Taylor's  series. — Heretofore  any  algebraic  expression  contain- 
ing X  has  been  spoken  of  as  a  function  of  x.  Thus  ex  -\-  e  is 
a  function  of  x,  c  and  e  being  costants.  Also  such  expressions 
as  e^a;+/i)^  sin(x  +  h),  tan  (a;  -f  h)  are  functions  of  x.  If,  however, 
a:  is  a  constant  and  h  a  variable  we  would  think  of  these  expres- 
sions as  functions  of  h. 

Let  y  be  any  function  whatever  of  {y  +  h),  let  a;  be  a 
constant  and  h  a  variable,  and  let  it  be  understood  that  y  and 
all  of  its  derivatives  are  finite  and  continuous.  Then  y  may 
be  expanded  by  Maclaurin's  theorem,  giving: 

y  [any  function  of  (x  +  h)]  -=A+Bh-\-  ~Ch^  +  ^D¥  +  •  •  •  (14) 

where  A,  B,  C,  D,  etc.,   are  constants,  as  follows: 

A   is  the  value  of    y     when  h  =  0 


158  CALCULUS. 

dv 
B  is  the  value  of  -4    when  ^  =  0 
an 

C  is  the  value  of   -r^  when  h  =  0 
etc.,  etc. 

Equation  (14)  is  sometimes  called  Taylor^s  series,  but  it  is 
identical  to  Maclaurin's  series  and  to  give  it  a  separate  name  is 
to  create  a  false  distinction. 

89.  Examples,     (a)  Expansion  of  e*. — The  successive  deriva- 
tives of  e*  are  as  follows: 

y  =  e'  which  is  equal  to  1  when  x  =  0 

dxi 

-^  =  e^  which  is  equal  to  1  when  a;  =  0 

j\  =  e*  which  is  equal  to  1  when  a;  =  0 

etc.,  etc. 

Therefore    A  =  B  =  C  =  D  =  etc.  =  1,    and  equation  (1)  of 
Art.  88  gives: 

e'  =  1  +  X  -^-^x^  +  j^  +  ^x'  -^  etc.  (1) 

(6)  Expansion  of  sin  x. — The  successive  derivatives  of  sin  x 
are  as  follows: 

2/  =  sin  a;  which  is  equal  to     0  when  re  =  0 

3^  =  cos  a;  which  is  equal  to     1   when  a;  =  0 
ax 

-t5  =  —  sin  a;  which  is  equal  to    0  when  a;  =  0 

-M  =  —  cos  a;  which  is  equal  to  —  1   when  a;  =  0 

t4  =  sin  a;  which  is  equal  to    0  when  x  =  0 
ax* 

and  so  on  in  endless  repetition. 


EXPANSIONS  IN  SERIES.  159 

Therefore    A  =  0,  5  =  1,  C  =  0,  D  =  ~  1,  ^  =  0,    etc.,  and 
equation  (1)  of  Art.  88  gives: 

/>t3  /v»5  /v»7  /v»v 

Binx  =  x-j-  +  j--j-  +  -^  etc.  (2) 

(c)  Expansion  of  cos  a;. — The  successive  derivatives  of  cos  x 
are  as  follows: 

y  =  cos  X  which  is  equal  to     1   when  x  =  0 

-^  =  —  sin  X  which  is  equal  to    0  when  x  =  0 

— I  =  —  cos  X  which  is  equal  to  —  1   when  x  =  0 

-^  =  sin  ic  which  is  equal  to    0  when  a;  =  0 

-T-^  =  cos  X  which  is  equal  to     1   when  x  =  0 

and  so  on  in  endless  repetition. 

Therefore    A  =  1,  5  =  0,  C  =  -  1,  D  =  0,  ^  =  1,  etc.,  and 
equation  (1)  of  Art.  88  gives: 

/ji»2  /v«4  /J.6  /»8 

COS  X  =  1  -^  +  ^  -  y  +  -|g-  etc.  (3) 

90.  Maclaurin*s  theorem  applied  to  a  function  of  two  inde- 
pendent variables. — If  u  is  a  function  of  x  and  y  which  is 
itself  finite  and  continuous,  and  if  all  of  its  derivatives  (partial 
derivatives)  are  finite  and  continuous,  then: 

u  =  A 
+  B.X  +  Byy 

+  i(C..x2  +  2C.yXy  +  Cyyy') 

+    ||-(i>xXXa^    +    SD,,yX^y    +    3D,yyXy^    +    Dyyyf)  (1) 

etc.        etc.        etc. 


160  CALCULUS. 

Where  A     is  the  value  of  u  when  x  and  y  are  both  zero. 

Bx    is  the  value  of  -j-   when    x    and    y    are  both  zero. 

d/li 

By    is  the  value  of  -r-    when    x    and    y    are  both  zero. 

Cxx  is  the  value  of  ^J  when   a:   and   ?/    are  both  zero. 

Cxy  is  the  value  of  ,    ,    or  -r-^  when  x  and  2/  are  both 

zero. 

Cyy  is  the  value  of  -r-y  when   a;   and   y   are  both  zero. 

etc.  etc.  etc. 

The  proof  of  Maclaurin's  theorem  as  applied  to  a  function  of  two  or  more 
variables  is  essentially  identical  to  the  proof  of  the  theorem  as  applied  to  a 
function  of  one  variable.  To  enable  one  to  appreciate  the  modified  character 
of  argument  and  especially  as  an  interesting  example  of  partial  integration, 
let  us  consider  what  we  may  call  the  second  approximation,  that  is  the  expres- 
sion we  get  for  u  on  the  assumption  that  the  respective  second  derivatives 
are  everywhere  constant  and  equal  to  Cxx,  Cxy  and  Cyy  (their  respective 
values  when  a;  =  0  and  j/  =  0)   as  follows: 

dhi       ^ 


Cxy  (3) 


dhi 
dxdy 

|S  =  C.  (4, 


Integrating  equation  (2)  with  respect  to  x  and  equation  (3)  with  respect  to  y 
we  have 

T-  =  CxxX  +  any  function  of  y  (5) 

and 

T-  =  CxyV  +  any  function  of  x  (6) 

Now  -T-  is  equal  to  Bx  when  x  and  y  are  both  equal  to  zero.    Therefore 

the  constant  term  in  "any  function  of  y"  in  (5)  is  equal  to  Bx,  and  the  constant 
term  in  "any  function  of    a;"    in  (6)  is  equal  to    Bx.    Therefore,  using  the 


EXPANSIONS  IN  SERIES.  161 

expression  "vanishing  function  of  ?/"  for  a  function  of  y  which  is  equal  to 
zero  when  y  =  0,   equations  (5)  and  (6)  become: 

-3-   =  CxxX  +  (a  vanishing  function  of  y)  +  Bx  (7) 

and 

;t-  '=  Cxyy  -\-  {Q'  vanishing  function  of  x)  -\-  Bx  (8) 

But  these  two  expressions  must  be  identical,  consequently  the  "vanishing 
function  of  ?/"  in  (7)  must  be  dyy  and  the  "vanishing  function  of  x^'  in 
(8)  must  be  CxxX.    Therefore  both  of  these  equations  reduce  to 

^  =  C.x  +  C.,7/  +  jB.  (9) 

In  a  similar  manner  we  may  integrate  equation  (4)  with  respect  to    y 
and  equation  (3)  with  respect  to  x,   and  get  the  equation: 

^=Cyyy+CxyX-\-By  (10) 

Now  equations  (9)  and  (10)  may  be  integrated,  giving: 

u  =  \CxxX^  +  Cxyxy  +  BxX  +  any  function  of  y  (11) 

and 

w  =  ICyyy^  +  CxyXy  +  Byy  +  any  function  of  x  (12) 

But  u  =  A  when  x  and  y  are  both  equal  to  zero,  therefore  "  any  function  of 
?/"  in  (11)  must  be  ("a  vanishing  function  of  ^")  +  -4,  and  likewise,  "any 
function  of  x"  in  (12)  must  be  ("a  vanishing  function  of  re")  -{-A.  Further- 
more, equations  (11)  and  (12)  must  be  identical  so  that  the  "vanishing  func- 
tion of  1/"  in  (11)  must  be  }4Cyyy^  +  Byy,  and  the  "vanishing  function  of 
x"  in  (12)  must  be  }4CxxX^  +  BxX.  Therefore  equations  (11)  and  (12)  both 
reduce  to: 

u  =  A 

+  BxX  +  Byy 
+  liCxxX^  +  2Cxyxy  +  Cyyy^)  (13) 

which  is  the  approximate  value  of  u  as  derived  from  equations  (2),  (3)  and  (4). 

PROBLEMS. 
1.  Expand      log(l  +  x)      by   Maclaurin^s   theorem.     Ans. 

/>»2  ^3  /v»4  /v*5 

log(l+a;)  =  x-2+3--4  +  5 

12 


162  CALCULUS. 

Note. — The  function  log  x  and  all  of  its  derivatives  become  infinite  for 
a;  —  0,  and  therefore  log  x  cannot  be  expanded  in  powers  of  x. 

2 .  Using  the  answer  to  problem  1  make  an  attempt  to  calculate 
the  logarithm  of  0  by  placing  x  =  —  1  and  adding  together  a 
number  of  terms  of  the  series. 

Note. — The  series  obtained  by  Maclaurin's  theorem  for  log  (1  +  x)  cannot 
be  used  for  values  of  x  which  lead  up  to  or  beyond  a  point  where  log  (1  +  x) 
or  any  of  its  derivatives  become  discontinuous  or  infinite. 

3.  Expand  cos(a;  +  h)  in  a  series  of  ascending  powers  of  h, 
Ans. 

cos  (a;  +  ^)  =  cos  a;  —  /i  sin  x  —  i^cos  x  +  "to"  sin  a;  +  •  •  • 

4.  Expand  y  =  tan  x  by  Maclaurin's  theorem.    Ans. 

,  ,  x^  ,  2x^  , 

tana:  =  a;+3-+-j5-+  ••• 

Note. — ^When  x  =  -  the  given  function  and  its  derivatives  become  infinite. 
Therefore  the  series  found  in  answer  to  this  problem  does  not  give  the  value  of 
tan  X  for  values  of  x  equal  to  or  greater  than  ^. 

91.  Demoivre*s  Theorem. — An  important  algebraic  identity, 
which  was  discovered  by  Demoivre,  is  expressed  by  the  equation: 

e'*  =  cos  a;  +  j  sin  X  (1) 

where  e  is  the  Napierian  base  and  j  =  V—  1.*  This  relation 
is  known  as  Demoivre^s  theorem,  and  it  is  very  useful  in  certain 
transformations  for  purposes  of  integration,  and  it  is  also  useful 
in  the  theory  of  alternating  currents. 

To  establish  equation  (1)  write  jx  for  x  in  equation  (1)  of 
Art.  89,  remembering  that  f  =  —  1,  f  —  —  jj  j^  =  +  1,  etc., 
and  we  have: 

-I       x^  .  x*     x^  .     .      ,    .  /        a^      x^      ^'  I     i.   \  /o\ 

*  It  is  customary  in  the  theory  of  alternating  currents  to  use  i  for  electric 
current,  and  j  is  used  for   V  —  1* 


EXPANSIONS  IN  SERIES.  163 

But  the  real  terms  in  this  series  give  a  series  identical  to  equation 
(3)  of  Art.  89,  and  the  imaginary  terms  give  a  series  identical  to 
equation  (2)  of  Art.  89.  Therefore  from  equation  (2)  we  get 
g/^  =  cos  X  +  j  sin  X. 

Definition  of  complex  quantity.  Geometric  representation  of 
complex  quantity. — Any  expression  like  a-\-h  V  —  1,  which  is 
part  real  and  part  imaginary  is  called  a  complex  quantity.  Thus 
the  right-hand  member  of  equation  (1)  is  a  complex  quantity; 
and  of  course  e^'*  is  a  complex 
quantity  because  equation  (1) 
shows  that  e^'*  is  part  real  (cos  x) 
and  part  imaginary   (j  sin  x). 

The   accepted    method  of    rep- 
resenting a  complex  quantity  geo- 
metrically  is   shown   in   Fig.    80.        '  pjg  go. 
The  vector  E  is  the  complex  quan- 
tity,  its   a:-component  is    thought  of    as   a   real   quantity    a, 
its    ^/-component  is  thought  of  as  the  imaginary  quantity    jh, 
and  the  vector  is  the  sum  of  its  two  components.     That  is 

E  =  a+jh* 

*  This  algebraic  expression  of  a  vector  as  a  complex  quantity  is  one  aspect 
of  the  important  use  of  complex  quantity  in  the  theory  of  alternating  currents. 
See  Franklin  and  Esty's  Elements  of  Electrical  Engineering^  Vol.  II,  Chapter  V; 
The  Macmillan  Co.,  New  York,  1908. 

Another  aspect  of  the  use  of  complex  quantity  in  the  theory  of  alternating 
currents  is  exhibited  in  Chapter  VII  of  this  treatise  where  the  fundamental 
differential  equations  of  alternating  currents  are  integrated  with  the  help  of 
transformations  involving  the  use  of  complex  quantity. 

The  practical  use  of  complex  quantity  in  alternating-current  theory  is  due 
chiefly  to  C.  P.  Steinmetz;  but  the  use  of  complex  quantity  in  the  solution  of 
linear  differential  equations  as  explained  in  Chapter  VII  contains  everything 
that  is  now  known  of  the  use  of  complex  quantity  in  alternating-current 
theory  in  a  manner  which  is  self-evident,  and  the  use  of  complex  quantity  as 
exemplified  in  Chapter  VII  is  much  older  than  electrical  engineering. 


164  CALCULUS. 

92.  Euler*s  expressions  for  sin  x  and  cos  x. — Many  differential 
expressions  can  be  reduced  to  simple  recognizable  forms  for 
purposes  of  integration  with  the  help  of  Demoivre's  theorem, 
and  it  is  in  some  cases  convenient  to  modify  equation  (1)  of 
Art.  91  so  as  to  express  sin  x  and  cos  x  in  terms  of  exponentials 
Such  expressions  are  due  to  Euler  and  they  are; 

sm  X  =  — ^ —  (1) 

and 

cos  X  = » (2) 

The  use  of  these  equations  is  exemplified  below.  They  are 
derived  as  follows:  From  Demoivre's  theorem  we  have: 

g;*  =  cos  a;  +  3  sin  x  (3) 

Write    —  a;   for  a;   in  this  expression,  remembering  that 

cos(—  x)  =  cos  a; 
and  that 

sin(—  x)  =  —  sin  x 
and  we  have: 

g-yx  =  cos  X  —  j  sin  x  (4) 

Subtracting  equation  (3)  from  equation  (4)  member  from  member 
we  get  equation  (1),  and  adding  equations  (3)  and  (4)  member 
to  member  we  get  equation  (2). 

93.  Example  showing  use  of  Euler's  equations. — Consider  the 
function : 

z  =  sin  mx  cos  nx  (1) 

To  find  the  average  value  of  this  function  between  x  =  0  and 
a;  =  27r  it  is  necessary  to  integrate  z-dx  between  the  Hmits 
x  =  0  and  a;  =  27r  as  explained  in  Art.  73.  Now  the  function 
given  in  equation  (1)  is  a  function  whose  average  value  is  of 
fundamental  importance  in  connection  with  Fourier's  theorem 
(see  Chapter  VIII)  and  therefore  it  is  important  to  be  able  to 


EXPANSIONS  IN  SERIES.  165 

reduce  the  differential  expression  sin  mx  cos  nx  -dx  to  a  combina- 
tion of  fundamental  forms  which  can  be  found  in  Class  A  of  the 
table  of  integrals  in  appendix  B.     This  reduction  may  be  easily 
made  with  the  help  of  Euler's  equations  as  follows: 
It  is  required  to  integrate  the  differential  equation: 

dy  =  sin  mx  cos  nx  •  dx  (2) 

Writing  mx  for  x  in  Euler's  expression  for  sin  x  we  get  an 
expression  for  sin  mx,  and  writing  nx  for  x  in  Euler's  expression 
for  cos  X  we  get  an  expression  for  cos  nx.  Substituting  these 
expressions  for   sin  mx   and  cos  nx   in  equation  (2),  we  get: 

dy  =  ^.  e^"("'+">^-rfa;  +  ^.e''^"'-''^'-"dx 

-  }^  e-i(»»+")=^  'dx-^.  e-J («-">^  •  dx     (3) 

Each  term  in  the  second  member  of  this  equation  is  of  the  form 
ae^^-dx    of  which  the  integral  (ignoring  constant  of  integration) 

is   r  •  &^.     Thus,  in  case  of  the  first  term   a  =  —.    and 

5  =  2{^  +  n) 

so  that  the  integral  of  the  first  term  is 

1 


4(m  +  n) 


e' 


{m-\-n)x 


Proceeding  in  a  similar  manner  with  each  term  and  arranging 
the  result  systematically  we  get: 


^  "  4(m  +  n)  ^^  ^  ^ 


4(m  —  n) 


Fg— y(m-n)x  _  g;(m— n)il        /^\ 


And  this  expression  can  be  easily  reduced  to  form  24  in  the 
table  of  integrals  by  using  Euler's  equations. 


166  CALCULUS. 

PROBLEMS. 

1.  Reduce  dy  =  sm^x-dx  to  a  combination  of  standard 
forms  as  given  in  the  table  of  integrals.     Ans. 

dy  ^  h-.  (e~'"'*  -  e^'  +  Z&'  -  Se"'')  *dx 

2.  Reduce  dy  =  coB>^X'dx  to  a  combination  of  standard  forms. 
Ans. 

dy  =  -hie^"^  +  e-''^  +  ^&^  +  ^e-^^  -\-^)'dx 

94.  Hjrperbolic  sines  and  cosines.* — In  the  solution  of  the 
differential  equation  of  the  alternating-current  transmission  linef 
the  expressions  (e*  —  e~^)  and  {e'  +  e"')  occur,  and  trans- 
mission line  calculations  are  facilitated  by  the  use  of  tables  giving 
the  values  of  (e*  —  e~^)  and  (e*  +  e~^)  for  various  values  of  x, 
and  of  course  the  discussion  of  such  calculations  is  simplified 
by  having  names  for    {e"  —  e~*)    and    {e'  +  e~').    Indeed  the 

expressions  ^ and  ^ are  related  to  the  equilateral 

hyperbola  in  the  same  way  that     ^. ( =  sin  x)      and 

^ ( =  COS  x)    are  related  to  the  circle.    Therefore  these 

expressions  are  called  the  hyperbolic  sine  of   x  (sinh  a;)  and  the 
*  Tables  giving  values  of  a,  h,  c  and  d  in  the  expressions 

cosh(aj  +  jy)  =  a  -^-jb 
and 

sinh(x  -^jy)  =  c  -\-jd 

for  various  values  of  x  and  y  are  published  in  a  supplement  to  the  General 
Electric  Review  for  May,  1910. 

t  The  simplest  discussion  of  this  subject  is  that  which  is  given  in  Franklin's 
Electric  Waves,  pages  141-153,  The  Macmillan  Co.,  New  York,  1909.  This 
discussion  is  essentially  complete,  although  no  mention  is  made  of  hyperbolic 
sines  and  cosines. 


EXPANSIONS  IN  SERIES.  167 

hyperbolic  cosine  of  x   (cosh  x)  respectively.    That  is: 


and 


sinh  X  = ^ (1) 


cosha;  =  ^^i^  (2) 


PROBLEMS. 

1 .  Differentiate  y  =  sinh  x» 

2.  Differentiate  y  =  cosh  x, 

3 .  Integrate  dy  =  sinh  x  •  dx. 

4.  Integrate  dy  =  coshx-dx. 

For  answers  see  forms  29  and  30  of  table  of  integrals  in 
Appendix  B. 


CHAPTER  VII. 
SOME  ORDINARY  DIFFERENTIAL  EQUATIONS. 

95.  Degree  and  order  of  a  differential  equation. — The  simple 
equation  az  -{-  h  =  0  is  said  to  be  linear  with  respect  to  z 
because  it  contains  no  power  of  z  higher  than  the  first  power. 
A  differential  equation  of  the  form: 

is  called  a  first  degree  or  linear  differential  equation  because  it 

contains  no  products  of  y,  ~   and  -—,  and  no  powers  of  2/,  ^, 

etc.,  higher  than  the  first.  The  coefficients  A,  B  and  C  in  a 
linear  differential  equation  may  contain  the  independent  vari- 
able X,  but  we  shall  confine  our  attention  in  this  chapter 
chiefly  to  linear  differential  equations  with  constant  coefficients. 

If  the  first  derivative,*  only,  appears  in  a  differential  equation, 
the  differential  equation  is  said  to  be  of  the  first  order.  If  the 
second  derivative  occurs  (with  or  without  the  first  derivative) 
the  differential  equation  is  said  to  be  of  the  second  order;  and 
so  on. 

96.  Ordinary  and  partial  differential  equations. — A  differential 
equation  expressing  the  law  of  growth  of  a  function  of  one  inde- 
pendent variable  is  called  an  ordinary  differential  equation. 
Many  examples  of  ordinary  differential  equations  are  given  in 
Chapters  I  and  V.     See  Art.  24  in  particular. 

*  The  terms  degree  and  order  apply  to  partial  differential  equations  as  well 
as  to  ordinary  differential  equations.  A  function  of  two  variables,  however, 
has  two  first  derivatives,  three  second  derivatives,  and  so  on  as  explained  in 
Art.  59.  Therefore  it  is  somewhat  misleading  to  speak  of  the  first  derivative, 
or  the  second  derivative  in  explaining  what  is  meant  by  a  differential  equation 
of  the  first  or  second  order. 

168 


i 


ORDINARY  DIFFERENTIAL  EQUATIONS.  169 

A  differential  equation  which  expresses  the  law  of  growth  of  a 
function  of  two  or  more  independent  variables  is  called  a  partial 
differential  equation.  Some  examples  of  partial  differential  equa- 
tions are  given  in  Chapter  V.     See  Arts.  60  and  90  in  particular. 

This  chapter  is  devoted  to  the  discussion  of  a  few  important 
ordinary  differential  equations,  and  several  important  partial 
differential  equations  are  discussed  in  Chapters  VIII  and  IX. 

97.  Pure  and  mixed  differential  equations.* — A  pure  differ- 
ential equation  contains  but  one  derivative  (the  first  or  any- 
higher  derivative)  and  does  not  contain  the  dependent  variable  y. 

Thus    dy  =  Qx-  dx,  -^  =  sin  x,  ^\  =  log  x    are  pure  differential 

equations. 

A  mixed  differential  equation  contains  more  than  one  derivative, 
or  it  contains  one  or  more  derivatives  and  the  dependent  variable 
y.     Thus 

are  mixed  differential  equations. 

Solution  of  pure  differential  equations. — A  pure  differential 
equation  can  be  integrated  by  looking  up  the  appropriate  form  in 
the  table  of  integrals.  This  is  exemplified  by  every  problem  in 
integration  heretofore  given  in  this  treatise.  In  the  case  of  a 
pure  differential  equation  of  the  second  or  third  order,  successive 
integrations  are  of  course  necessary.  For  example,  consider  the 
third  order  pure  differential  equation: 

Let  q  represent  ~,   then  equation  (1)  becomes: 

1  =  -'  (2) 

*This  and  the  following  articles  refer  primarily  to  ordinary  differential 
equations. 


170  CALCULUS. 

This  equation  can  be  integrated,  giving; 


9  =  3  =  4<^  +  C  (3) 


Let  p  represent   -p,   then  equation  (3)  becomes: 


%-ia^  +  C  (4) 

This  equation  can  be  integrated,  giving: 

P  =  ^  =  i\ax'  +  CX  +  D  (5) 

This  equation  can  be  integrated,  giving 

y  =  -hax^  -f  ^Cx^  +  Dx  +  E  (6) 

Solution  of  mixed  differential  equations.  Separation  of 
variables. — The  linear  differential  equation  (1)  of  Art.  95  is  of 
course  a  mixed  equation,  and  the  general  solution  of  this  equation 
with  constant  coefficients  is  given  in  a  subsequent  article.  We 
will  here  consider  one  or  two  examples  of  a  simple  transformation, 
called  the  separation  of  variables,  which  can  sometimes  be  used 
to  bring  a  mixed  differential  equation  into  a  form  which  can  be 
integrated  by  looking  up  appropriate  forms  in  the  table  of 
integrals. 

As  a  first  example  consider: 


This  equation  reduces  to: 


1  =  ^  (^) 


^  =  x'^'dx  (8) 

y 


and  this  equation  can  be  integrated  with  the  help  of  forms  1  and 
2  of  the  table  of  integrals,  giving 

log2/  =  |x3  +  C  (9) 


ORDINARY  DIFFERENTIAL  EQUATIONS.  171 

As  a  second  example  consider 

dij 
Let  p  represent   -^   and  this  equation  becomes 

g  =  x«p  (12) 

and  the  integral  of  this  equation,  according  to  equation  (9),  is: 

\ogv  =  \:x?  -\-  C 
or 

P  =  I  =  ei^'^"  (13) 

This  is  a  pure  differential  equation  and  its  integral  can  be  found 
by  looking  up  the  appropriate  forms  in  the  table  of  integrals. 

PROBLEMS. 
Solve  the  following  differential  equations: 

1.  x3  ^  =  2.    Ans.  2/  =  log  a;  +  Cx"  ■\- Dx -^  E, 

2.  ^  =  xe',    Ans.  y  =  {x  -  2)e'  +  Cx  +  D. 

3.  ^  =  27  sin^x.    Ans.  y  =  21  cos  x-  cos^  x  -\-Cx^  +Dx  -{-E, 

4.  1  -  2/  +  (1  +  a;)  ^  =  0.    Ans.  log  \-^  =  C. 

ax  1  ~~  2/ 

5.  1  -  21/  =  3^.     Ans.   C(l  -  2y)  =  e"?. 

6.  sin  2/-cix  +  a;  cos  i/-(f?/  =  0.    Ans.  xsiny  =  C. 

7.  sin  a;  cos  y-dx  —  cos  x  sin  y  -  dy  =  0.     Ans.  cos  y  =  C  cos  a;. 

8.  fa  +  a;i/)rfa;  +  (a;  —  xy)dy  =  0.    Ans.  logxy  =  C  —  x  +  y. 

9.  Vl  —  2/^  •  (ia;  +  V 1  —  a;^  •  (^2/  =  0.    Ans.  sin""^  x-\-  sin"^  y=C. 

^°-S  +  .-|  =  "-    Ans.Clogx  =  .  +  D. 


172  CALCULUS. 


11.  ^  +  (^f^y  +  1=0.     Ans.  y  =  log  cos  {x  -  C)  +  D. 


12.(l+.^)g  +  .|  +  ax  =  0. 


Ans.  y  =  D  —  ax  -\-  C  log  [x  -\-  Vl  +  x^]. 

98.  The  general  solution  and  particular  solutions  of  a  differ- 
ential equation. — The  general  solution  of  a  differential  equation 
is  an  expression  for  y  (the  dependent  variable)  in  terms  of  x 
(the  independent  variable)  which  includes  every  possible*  func- 
tion which  satisfies  the  differential  equation.  Thus  equation  (6) 
of  Art.  97  is  the  general  solution  of  equation  (1)  of  that  article. 
Concerning  the  three  constants  of  integration  it  is  evident  that 
E  is  the  value  of  y  when  a;  =  0.  Also  it  is  evident  from  equa- 
tion (5)  of  Art.  97  that  D  is  the  value  of  -^   when  x  =  0,  and 

it  is  evident  from  equation  (3)  that  C  is  the  value  of  — g  when 

X  =  0. 

The  general  solution  of  a  differential  equation  of  the  nth.  order 
contains  n  undetermined  constants  of  integration. 

When  one  or  more  of  the  constants  of  integration  in  the  general 

solution  of  a  differential  equation  have  particular  values  assigned 

to  them  we  have  what  is  called  a   particular   solution  of  the 

differential  equation.     For  example,  let  C  =  2,  let  D  =  0  and 

let  E  =  0  in  equation  (6)  of  Art.  97,  then  the  equation  becomes 

ax^ 
y  =  j^  -\-  x^    which  is  a  particular  solution  of  equation  (1)  of 

Art.  97. 

99.  Discussion  of  the  first  order  linear  differential  equation 
with  constant  coefficients. — Consider  the  differential  equation: 

y  +  Ag  =  o  (1) 

*  This  statement  is  subject  to  some  qualification  because  of  what  is  called 
the  singular  solution.  See  Johnson's  Ordinary  and  Partial  Differential  Equa- 
tions, page  43,  John  Wiley  &  Sons,  New  York,  1890. 


ORDINARY  DIFFERENTIAL  EQUATIONS.  173 

The  most  obvious  method  for  solving  this  differential  equation 
is  to  "separate  the  variables"  as  explained  in  Art.  97.  Thus 
equation  (1)  is  easily  reduced  to: 

A^=-dx  (2) 

y  ^  ^ 

and  this  equation  can  be  integrated  by  looking  up  appropriate 
forms  in  the  table  of  integrals,  giving: 

Alogy  =  -x-VC  (3) 

It  is  desirable,  however,  to  solve  equation  (1)  by  the  following 
method  because  the  method  applies  to  a  linear  differential  equa- 
tion of  any  order  when  the  coefficients  are  constants. 
Let 

y  =  Ce*^  (4) 

where  C  and  k  are  constants,  and  e  is  the  Napierian  base. 
Then: 

Substituting  the  values  of  y  and  ~-  from  (4)  and  (5)  in  equation 

(1),  we  have: 

Ce^^  +  AkCe^'  =  0  (6) 

whence  by  cancellation  we  have: 

1  +  AA;  =  0 


or 

t  =  — 
A 


k  =  -1  (7) 


Therefore  if  h  in  equation  (4)  has  the  value  —  -j  then  equation 
(4)  satisfies  equation  (1);  that  is,  equation  (4)  is  a  solution  of 
(1)  if   k  =  —  -T-',   indeed  equation  (4)  is  the  general  solution  of 


174  CALCULUS. 

(1)  because  it  contains  one  undetermined  constant     C,     the 
constant  of  integration. 

100.  The  principle  of  superposition. — A  principle  of  extremely 
wide  application  in  physics  is  the  so-called  principle  of  super-posi- 
tion.   From  the  physical  point 
window  No,  1  window  No.  2    of  view  a  general  statement  of 

this  principle  is  scarcely  possi- 
ble and  therefore  the  following 
examples  must  suffice:  (a)  A 
person  at  A  (Fig.  81)  can  see 
window  No.  1  and  another  per- 
son at  B  can  see  window  No.  2 
Fig.  8L  at  the  same  time.     This  means 

that  two  beams  of  light  a  and  h 
can  travel  through  the  same  region  at  the  same  time  and  not 
get  tangled  up  together  as  it  were,  each  beam  behaving  as  if  it 
were  traveling  through  the  region  alone.  (6)  Two  sounds  can 
travel  through  the  same  body  of  air  simultaneously,  each  sound 
behaving  as  if  it  were  traveling  through  the  body  of  air  alone, 
(c)  Two  systems  of  water  waves  can  travel  over  the  same  part 
of  a  pond  simultaneously,  each  system  behaving  as  if  the  other 
were  not  present,  {d)  Two  messages*  can  travel  over  a  tele- 
graph wire  at  the  same  time  and  not  get  mixed  up.  (e)  Two 
forces  F  and  G  exerted  simultaneously  upon  an  elastic  structure 
produce  an  effect  which  is  the  sum  of  the  effects  which  would  be 
produced  by  the  forces  separately,  provided  the  sum  of  the  forces 
does  not  exceed  the  elastic  limit  of  the  structure;  therefore  each 
force  may  be  thought  of  as  producing  the  same  effect  that  it 
would  produce  if  it  were  acting  alone 

All  of  the  effects  in  physics  which  are  superposable — and  this 

*  Indeed  any  number  of  messages  can  travel  over  a  telegraph  wire  in 
either  direction  or  in  both  directions  simultaneously.  The  only  limiting 
feature  in  multiplex  telegraphy,  when  line-loss  is  negligible,  is  in  the  design  of 
the  sending  and  receiving  apparatus;  and  the  same  is  true  in  wireless  teleg- 
raphy.   In  each  of  the  above  examples  the  word  two  means  iwo  or  more. 


ORDINARY  DIFFERENTIAL  EQUATIONS.  175 

includes  the  greater  part  of  the  effects  in  mechanics,  heat, 
electricity  and  magnetism,  light  and  sound,  and  a  great  many- 
effects  in  chemistry — are  expressible  in  terms  of  linear  differential 
equations  with  constant  coefficients,  and  the  principle  of  super- 
position may  be  thought  of  as  a  property  of  such  a  differential 
equation  as  follows .  If  y  is  a  function  of  x  which  satisfies  a 
linear  differential  equation  with  constant  coefficients,  and  if  z  is 
another  function  of  x  which  satisfies  the  same  differential  equation, 
then    (y  +  z)    is  a  function  of  x   which  satisfies  the  equation* 

Proof. — Let  the  given  linear  differential  equation  be: 

,    A  du   .    „  d?u   .  f.  /^v 

If  a  function  y  satisfies  this  equation,  then: 

If  another  function  z  satisfies  equation  (1),  then: 

Now  ^^y  +  ^'>  =  ^  +  ^  and  '^^^  +  ^>  _  ^  .  ^  There- 
^""^        dx  dx^  dx  *""*        dx^  dx'  +  dx»-     "^"^^ 

fore,  adding  equations  (2)  and  (3),  we  get: 

(,  +  .)+A'?(^  +  B^J-i)+...=0  (4) 

But  this  is  the  same  form  as  equation  (1)  which  shows  that  the 
function    {y  +  z)    satisfies  (1). 

*  This  proposition  is  true  for  a  partial  linear  differential  equation  also. 
Indeed  most  of  the  superposable  effects  in  physics  are  expressible  in  terms  of 
partial  linear  differential  equations  with  constant  coefl&cients.  Examples 
are  given  in  Chapters  VIII  and  IX. 


176  CALCULUS. 

101.  Discussion  of  the  second  order  linear  differential  equa- 
tion with  constant  coefficients. — Consider  the  differential  equa- 
tion: 

It  is  not  possible  to  "separate  the  variables"  in  this  differential 
equation  and  therefore  the  second  method  of  Art.  99  must  be 
used.    Therefore  let: 

y  =  Ce**  (2) 

then 

|  =  iCe*.  (3) 

and 

Substituting  these  values  of  y,  -^   and  -r-^  in  equation  (1)  and 

cancelling  the  common  factor  Ce**,   we  have : 

1  4-  A/b  +  M2  =  0  f\    ^  (5) 

whence 


Therefore  using  a  for  one  of  these  values  of  k  and  using  j8  for 
the  other  value  of  A;,  we  get  two  solutions  of  (1),  namely: 

w  =  Ce'^  (7) 

and 

z  =  De^'  (8) 

But  according  to  Art.  100  the  sum    (w  +  z)    is  also  a  solution. 
Therefore  using   y   for    (w  -{-  z),    we  have  as  a  solution  of  (1): 

2/  =  C6-  +  i)e^-         iW^-v-ii--^      (9) 

and  this  is  the  general  solution  of  (1)  because  n  contains  the  two 
undetermined  constants   C  and  D. 


ORDINARY  DIFFERENTIAL  EQUATIONS.  177 

102.  The  starting  of  a  boat. — At  a  certain  instant    (t  =  0)    a 

constant  force    E    begins  to  act  on  a  boat,  and  it  is  desired  to 

find  an  expression  for  the  increasing  velocity    i    of  the  boat. 

At  very  low  speeds  the  backward  drag  or  friction  of  the  water  on 

a  boat  is  proportional  to  the  velocity  of  the  boat.     Let  us  assume 

that  this  proportional  relation  is  exact,  then  the  frictional  drag 

of  the  water  on  a  boat  is  equal  to  Ri,  where  i  is  the  velocity  of 

the  boat  and    R    is  sl  constant  which  depends  on  the  shape  and 

size  of  the  boat.     Therefore  the  net  accelerating  force  acting  on 

the  boat  is     E  —  Ri.     This  force  is  equal*  to  the  product  of 

di 
the  mass  L  of  the  boat  and  the  acceleration  -r;.     Therefore  we 

at 

have: 

L%^E-Ri  (1) 

To  get  this  equation  into  the  standard  form  of  a  linear  differential 
equation,  let 

y  =  E-Ri  (2) 

Then 

^  =  _  p^ 
dt  ^dt 

and  equation  (1)  becomes: 

R     dt      ^ 
or 

^  +  1-1  =  0  (^) 

The  general  solution  of  this  equation  is  given  in  Art.  99,  but 
it  is  worth  while  to  work  it  out  anew,  as  follows:  Let 


y  =  Ce^«  (4) 

*  If  force  is  expressed  in  poundals  (or  dynes),  mass  in  pounds  (or  grams),  and 
acceleration  in  feet  per  second  per  second  (or  centimeters  per  second  per 
second). 
13 


178  CALCULUS, 

then 

dt 


^^  =  A;Ce«  (5) 


dv 
Substituting  these  values  of    y    and    -^    in  equation  (3)  and 

cancelling  out  the  factor  Ce*',   we  have: 

80  that 

"  L 

Therefore  equation  (4)  becomes: 

y  =  Ce"^"' 
or,  substituting  E  —  Ri  for  yj  we  have: 


(6) 


E-  Ri  =  Ce  ^ 


(7) 


Now  i  —  Q  when  t  =  0.     Therefore  placing  this  pair  of  values 
in  equation  (7)  we  have: 

E  ^C  (8) 

so  that  the  constant  of  integration  is  determined,  and  equation 
(7)  becomes: 

E  -  Ri  ^  Ee  ^ 
or 


R       r" 


|-|e-^-  (9) 


This  equation  expresses  the  value  at  each  instant  of  the  increasing 
velocity  of  the  boat  as  a  function  of  the  elapsed  time  t. 

103.  The  stopping  of  a  boat. — A  boat  is  moving  at  velocity  I 
when  the  propelling  force  ceases  to  act,  and  it  is  required  to  find 
an  expression  for  the  decreasing  velocity,    i,    of  the  boat  as  it 


ORDINARY  DIFFERENTIAL  EQUATIONS. 


179 


gradually  comes  to  rest.     In  this  case  the  only  force  acting  on 
the  boat  is  the  retarding  force  Ri,  and  in  this  case  the  accelera- 


di 
tion   T7  is  negative  so  that: 


or 


^  +  §•1  =  0 


The  general  solution  of  this  differential  equation  is: 


(1) 
(2) 

(3) 


But   i  =  /   when   t  =  0.     Therefore    C  =  /,    and  equation  (3) 
becomes: 


i  =  le 


(4) 


Remark. — The  notation  used  in  this  discussion  of  the  starting 
and  stopping  of  a  boat  is  the  standard  notation  used  in  electrical 
theory.     This  notation  is  used  here  and  in  the  following  articles 


35 
30 

25 
«20 














■ 

-5 

p- 

. 



^ 

-^ 

' 

^ 

>^ 

/ 

. 

/ 

grc 
R= 

wing 
'3  oh 

current 

ms     1 

R 

/ 

L  = 

0.04 
110 

hem 
polts 

V 

J_ 

3         4        5        6         7        8         9 

hundredths  of  a  second 

Fig.  82. 


180 


CALCULUS. 


because  the  mechanical  problems  discussed  in  this  and  the 
following  articles  are  strictly  analogous  to  certain  electrical 
problems  which  constitute  the  foundation  of  the  theory  of 
alternating  currents.  Thus  the  curve  in  Fig.  82  is  a  graphical 
representation  of  equation  (9)  of  Art.  102;  the  ordinates  of  this 
curve  represent  the  growing  values  of  the  current   i   in  a  circuit 


\- 

35 

K 

J^ 

^ 

•^b 

\ 

dei 
R  = 

:ayin 

-3  oh 

0.04 

367 

g  current 
img 

^20 

\ 

\ 

henry 
amperes 

\ 

\ 

lO 

\ 

\. 

S 



, 

• 

2 

3 

mndr 

edth 

5        ( 

8  of  < 

3 

I  sec 

1 
ond 

3        < 

? 

Fig.  83. 

of  resistance  R  and  inductance  L  after  a  battery  of  electro- 
motive force  E  is  connected  to  the  circuit.  The  curve  in  Fig. 
83  is  a  graphical  representation  of  equation  (4)  of  Art.  103;  the 
ordinates  of  this  curve  represent  the  decaying  values  of  the 
current  i  when  an  initial  current  of  /  amperes  is  left  to  die 
away  in  a  circuit  of  resistance   R   and  inductance  L. 


PROBLEMS. 
Solve  the  following  differential  equations. 

Ans.  y  =  Ce-'  +  De^'. 


-3-2-^^  =  0- 


^-  da? 


4y.     Ans.  y  =  Ce^'  +  De-^', 


"-  -^  ^^ 


ORDINARY  DIFFERENTIAL  EQUATIONS. 


181 


^-4^  +  3)/  =  0.    Ans.  y  =  Ce' +  D^'. 


<Py 


•n^^+^ 


)  = 


10 


dy 
dx 


'■%-4i^'t  =  '- 


Ans.  y  =  Ce^*  +  Be^'. 
Ans.  y  =  Ce^^  +  De^^  +  E. 


104.  Undamped  oscillations. — Figure  84  shows  a  weight  sus- 
pended by  a  spring,  and  the  weight  stands  in 
equilibrium  in  a  certain  position.     If  the  weight 
happens  to  be    q   feet  above  or  below  its  equilib- 
rium position  an  unbalanced  force  proportional 

to  q,  and  therefore  equal  to   y^*    times    q    acts 


upon  the  weight.  This  force  is  downwards  when 
q  is  upwards,  and  upwards  when  q  is  down- 
wards.    Therefore  the  force  is  equal  to    —  p  -q- 

This  force  accelerates  (or  retards)  the  weight, 
and  we  have 


Y  di  _ 


(1) 


spring 


Fig.  84. 


where  L  is  the  mass  in  pounds  of  the  attached 

di 
weight  in  Fig.  84,  and  -^  is  the  acceleration  (it  is 

a  retardation  when  it  is  negative)  of  the  attached 

weight.     But  the  velocity  i  of  the  weight  is  the 

rate  of  change  of  its  distance    q    from  the  equilibrium  position 

That  is 

*      dt 

so  that 

(U  _^ 

dt  ~  dt^  ^"^^ 

*  This  is  simply  a  proportionality  factor.  It  is  written  in  this  form  so  a 
to  conform  to  the  standard  notation  of  electrical  theory. 


(2) 


182  CALCULUS. 

Therefore  equation  (1)  becomes 


or 


T  ^  _  _q 
dt^  C 

.  +  §•9  =  0  (4) 


The  general  solution  of  this  second  order  linear  differential  equa- 
tion is  given  in  Art.  101,  but  it  is  worth  while  to  work  out  the 
solution  anew  as  follows: 
Let 

q  =  Me''*  (5) 

then 

^  =  /bM6**  (6) 


and 


kme"*  (7) 


Substituting  these  values  of    q    and    ■—    in  equation  (4)  and 
cancelling  the  common  factor   ilfe**,   we  have: 


1  +  ^'k''  =  0  (8) 


so  that 


k  =  yj 


-c  («) 


It  is  evident  that  k  is  imaginary  because  L  and  C  are  positive 
quantities.  Therefore  let  us  write  j<a  for  k  where  j  is  V  —  1,* 
then:  _ 

IL 


\C 


(9) 


*  It  is  customary  in  electrical  theory  to  use  j  for  V  —  I  because  the  letter 
t  is  used  for  electric  current. 


ORDINARY  DIFFERENTIAL  EQUATIONS.  183 

Now  of  course  w  may  be  either  positive  or  negative,  that  is,  k 
may  be    +  jco   or    —  jw.     Therefore 

V  =  Me+i'^*  (10) 

and 

z  =  Ne-i^'  (11) 

are  two  particular  solutions  of  equation  (4),  where  M  and  N 
are  undetermined  constants.  Therefore,  according  to  Art.  100, 
(v  +  z)  is  also  a  solution  of  (4).  Consequently,  writing  u  for 
{v  ■{•  z),   we  have  as  a  solution  of  (4) : 

u  =  Me+'"'  +  Ne-^"**  (12) 

and  this  is  the  general  solution  of  (4)  because  it  contains  the 
two  undetermined  constants   M   and   N. 

Now  from  Demoivre's  theorem  it  is  evident  that  u  as  given 
by  equation  (12)  is  complex  (part  real  and  part  imaginary),  and 
for  the  sake  of  generahty  M  and  N  may  be  thought  of  as 
complex  also.    That  is,  for  M  and  N  we  may  write: 

M  =  M'  -{-  jM''  (13) 

and 

iV  =  iV'  +  jN"  (14) 

Now  any  complex  equation  is  equivalent  to  two  simple  equations, 
namely  the  equation  which  expresses  the  equaUty  of  the  real 
parts  of  the  members  of  the  complex  equation,  and  the  equation 
which  expresses  the  equality  of  the  imaginary  parts  of  the  com- 
plex equation;  and  this  latter  equation  becomes  real  when  j 
is  cancelled  from  its  members.  Thus  equation  (12)  may  be 
broken  up  into  two  simple  equations  each  of  which  is  a  solution 
of  equation  (4). 

To  spHt  equation  (12)  up  into  two  equations  use  the  values 
for  M  and  N  from  equations  (13)  and  (14),  and  reduce  e+^"'*" 
and  e~''^'  by  means  of  Demoivre^s  theorem  as  explained  in  Art. 
91.     We  thus  get: 

r  -Vjs  =  {W  +  jM")  (cos  (at  +  j  sin  coO  /.  ^x 

+  {N'  +  jN")  (cos  co<  -  3  sin  coQ     ^  ^ 


184 


CALCULUS. 


where    r  +  js    has  been  written  for    q.    Separating  real  and 
imaginary  terms  in  equation  (15)  we  get: 


and 


r  =  (ilf '  +  N')  cos  a)t  -  {M"  -  N")  sin  co<  (16) 

8  =  (M"  +  N")  cos  oit  +  {M'  -  N')  sin  cot  (17) 

Now  in  either  of  these  equations  the  coefficient  of  cos  o)t  is  any 
constant,  and  the  coefficient  of  sin  cat  is  amj  constant.  There- 
fore, q  for  r  or  s,  both  of  these  equations  reduce  to: 


q  =  G  cos  cot  +  H  sin  cot 


(18) 


which  is  the  final  general  solution  of  equation  (4),    G    and    H 
being  undetermined  constants. 

In  alternating  current  theory  this  equation  is  used  in  a  slightly 
modified  form  as  follows: 

Let 

G  =  -  0  sin  0  (19) 


and 

H  =  Qcosd 

Then  equation  (18)  becomes: 

g  =  Q  sin  (co<  —  d) 


(20) 


(21) 


This  equation  is  represented  by  Fig.  85,  in  which  the  line    Q 

rotates  at  angular  velocity  co 
radians  per  second,  and  the 
value  of  q  at  each  instant  is 
represented  by  the  projection 
of   Q    on  the  fixed  line  ah. 

The  two  integration  con- 
stants Q  and  d  are  deter- 
mined by  the  initial  condi- 
tions as  follows:  For  example 
let  the  weight  be  started 
oscillating  by  pulling  it  up- 
wards  2   feet     (q  =  2   feet), 


axis  of  angle. 


ORDINARY  DIFFERENTIAL  EQUATIONS.  185 

holding  it  still  ( 37  =  0  1  and  releasing  it  at  the  instant  t  =  0. 
The  values  of  Q  and  6  are  then  determined  by  substituting  the 
values   37  =  0   and  i  =  0   in  the  equation 

~  =  ojQ  cos  (Q)t  —  6) 

and  by  substituting  the  values  q  =  2  and  t  =  0  inequation  (21). 

105.  Damped  oscillations. — The  weight  in  Fig.  84  is  acted 

upon  by  the  force    —  ^    when  it  is  at  a  distance    q    above  or 

below  its  equiHbrium  position  as  explained  in  Art.  104;  and, 
if  the  weight  moves  up  and  down  through  a  resisting  fluid,  the 
backward  drag  due  to  friction  will  be  approximately  proportional 

to  the  velocity    -^    of  the  weight.     Assuming  this  proportional 

relationship  to  be  exact,  the  backward  drag  of  friction  may  be 

written    R'-^,    and  since  it  is  always  opposed  to  the  velocity 

-^   it  must  be  written    —  R~,     Therefore  the  total  force  acting 

on  the  weight  is     —  ^  —  72  -^,     and  this  force  is  equal  to  the 

(Pa 
product  of  the  mass   L   of  the  weight  and  the  acceleration   — 

of  the  weight.     Therefore  we  have: 

T^  -  -  1  -.  J?^ 
^dt^  ~       C       ^dt 
or 

3  +  Cijf  +  Lcg  =  0  (1) 

The  general  solution  of  this  equation  may  be  found  by  the 
method  used  in  Art.  104.     Reduced  to  simple  form  the  general 


186 


CALCULUS. 


solution  is; 


q  =  Qe-^  sin  {(at  -  B) 


(2) 


where  Q  and   B  are  undetermined  constants  of  integration  and 
where  a  and  w  are  written  for  the  following: 


a  — 


and 


2L 


\Z^~ 


4L2 


(3) 


(4) 


Equation  (2)  completely  expresses  the  motion  of  the  weight  in 
Fig.  84  when  the  motion  of  the  weight  is  opposed  by  friction  as 
above  explained.  Indeed  equation  (2)  expresses  a  kind  of 
harmonic  motion  in  which  the  amplitude  (Qe~*0  decreases 
continually. 


q-axU 


'"^^^^^^^xponentiai  curve 


Fig.  86. 

The  varying  value  of  q  as  expressed  by  equation  (2)  is  repre- 
sented by  the  ordinates  of  the  wavy  curve  in  Fig.  86  for  the  case 


in  which    B  =  — 


The  dotted  curves  are  exponential  curves 


ORDINARY  DIFFERENTIAL  EQUATIONS. 


187 


of  which  the  equations  are  g  =  ±  Qe~^,  Another  graphical 
representation  of  equation  (2)  is  shown  in  Fig.  87  where  q  is 
the  projection  on  the  fixed  line  ah  of  a  line  Qe"**  which  rotates 


01 

I     ^logarithmic  spiral 


Fig.  87. 


at  a  constant  angular  velocity  of  w  radians  per  second  and 
grows  continually  shorter  and  shorter.  The  end  of  the  Une 
Qe~**  describes  the  dotted  curve  which  is  called  an  exponential 
spiral. 

PROBLEMS. 

Solve  the  following  differential  equations: 


+  42/  =-  0.     Ans.  2/  =  C  sin  2a;  +  D  cos  2x. 


2.  ^  -  2  ^  +  5i/  =  0.     Ans.  y  =  e-(C  sin  2x  +  D  cos  2x). 

3.  ^  -  8  ^  +  25?/  =  0.     Ans.  y  =  e^(C  sin  Zx  +  D  cos  Sx). 

4.  ^  -  24  ^  +  1692/  =  0.     Ans.  y  =  e^^x^e  gi^  5x-{-D  cos  5a;). 


188 


CALCULUS. 


5.  ^H-^  +  2/  =  0.     Ans.  y  =  e-^'(c 8m"fx+D co8^x\ 

106.  Forced  or  maintained  oscillations. — The  weight  in  Fig. 
84  is  acted  upon  by  the  force    —  ^  due  to  the  altered  stretch  of 

the  spring,  and  by  the  force    ~  ^  jT    due  to  the  frictional  drag 

of  the  fluid  in  which  the  weight  moves  up  and  down.  Let  us 
suppose  that  an  outside  periodic  force,  E  sin  pt,  acts  upon  the 
weight.     Then  the  total   force   acting  on  the  weight  will  be 

and  this  will  be  equal  to  the  product 


l-Bf^+E^npt, 


of  mass  times  acceleration 


m 


Therefore  we  will  have: 


q  +  CR^  +  LC^  =  CEsmpt 


(1) 


circuit 


This  is  the  differential  equation  of  motion  of  the  weight  under 

the  assumed  conditions,  and 
this  identical  differential  equa- 
tion expresses  the  mode  of  vari- 
ation of  the  charge  q  in  the 
condenser  C  in  Fig.  88  when 
an  alternator  of  which  the 
electromotive  force  is  E  sin  pt 
delivers  electric  current  to  a  cir- 
n  cuit  of  resistance  R  and  induc- 
tance L,  the  circuit  containing 
a  condenser  of  capacity  C. 
The  charge  q  in  the  condenser 
is  not  so    familiar  a    thing  as 


I  alterwUor 


circuit 

Fig.  88. 


the  current  i 


(■§) 


which  flows  in  the  circuit,  but  it  is  con- 
venient to  use  q  as  our  variable  instead  of  i. 


ORDINARY  DIFFERENTIAL  EQUATIONS.  189 

Everyone  who  has  had  any  experience  with  alternating-current 
phenomena  knows  that  in  general  there  are  two  recognizable 
alternating  currents  in  the  circuit  shown  in  Fig.  88,  namely,  (a) 
The  alternating  current  which  is  maintained  hy  the  alternator. 
This  alternating  current  is  of  the  same  frequency  as  the  electro- 
motive force  E  sin  yt  of  the  alternator,  and  (6)  A  decaying 
oscillatory  current,  as  if  the  condenser  had  been  initially  charged 
and  allowed  to  discharge  through  the  circuit.  This  decaying 
alternating  current  is  independent  of  the  alternator,  it  is  expressed 
by  equation  (2)  of  Art.  105,  and  its  frequency  is  determined  by 
equation  (4)  of  Art.  105. 

The  finding  of  the  general  solution  of  equation  (1)  is  accom- 
plished in  two  steps.  The  first  step  is  the  finding  of  a  so-called 
singular  solution  of  (1),  a  solution  which  expresses  the  maintained 
alternating  current  above  mentioned,  and  the  second  step  is  to 
add  to  this  singular  solution  the  general  solution  of  equation  (1) 
of  Art.  105  as  given  by  equation  (2)  of  Art.  105.  This  addition 
is  permissible  inasmuch  as  it  is  very  easy  to  show  that  v  +  z 
is  a  solution  of  equation  (1)  of  this  article  if  t;  is  a  solution,  and 
if  z   is  a  solution  of  equation  (1)  of  Art.  105. 

This  procedure  involves  elaborate  transformations  which  are 
unintelligible  to  the  beginner,  and  it  is  therefore  useless  to  carry 
it  out.  Fortunately,  however,  a  much  simpler  method  is  avail- 
able for  the  establishment  of  the  desired  result,  a  result  which  is 
the  foundation  of  the  more  important  part  of  the  theory  of 
alternating  currents.* 

*  See  pages  66-68  and  73-74  of  Franklin  and  Esty's  Elements  of  Electrical 
Engineering,  Vol.  II. 


CHAPTER  VIII. 


THE  PARTIAL  DIFFERENTIAL  EQUATION  OF  WAVE  MOTION. 

107.  Differential  equation  of  travel.  Equation  of  a  traveling 
curve. — ^When  a  point  is  stationary  its  abscissa  x  is  constant, 
and  when  a  point  travels  at  constant  velocity  along  the  x-axis  of 
reference  its  abscissa  increases  (or  decreases)  at  a  constant  rate 
so  that  X  =  kt  -\-  any  constant.  This  is  all  very  simple  but  the 
equation  of  a  traveling  curve  which  is  so  extensively  used  in  the 
theory  of  wave  motion  is  not  so  simple,  and  therefore  some  dis- 
cussion of  it  is  necessary. 

The  curve  cc  in  Fig.  89  is  stationary  with  respect  to  the  origin 
O'j   and  the  equation  of  the  curve  referred  to  the  origin  0'  is: 


y  =  Fix') 


(1) 


stationary 
y-axia 


^mooing 
^  y-axia 


1 


j  x-axU 


vt * X 

d X    —  —  —  - 


Fig.  89. 


But  the  origin  0'  and  the  curve  cc  are  both  assumed  to  be 
traveling  to  the  right  at  velocity  v,  so  that  the  abscissa  of  the 
moving  origin  0'  referred  to  the  stationary  origin  0  is  vt.  There- 
fore the  abscissa  x  of  any  point  on  the  curve  cc  referred  to  the 
stationary  origin  is  equal  to  x'  +  vt,  or  x'  =^  x  —  vt.  Therefore, 
substituting  x  —  vt  for  x'  in  equation  (1)  we  have 


y  -  F{x  —  vt) 
190 


(2) 


DIFFERENTIAL  EQUATION  OF  WAVE  MOTION.  191 

which  is  the  equation  of  the  traveling  curve    cc    referred  to  the 
stationary  origin  0. 
Similarly  it  may  be  shown  that 

y=f{x  +  vt)  (3) 

is  the  equation  of  a  curve  which  is  traveling  to  the  left  at  velocity  v. 
The  two  equations  (2)  and  (3)  satisfy  one  of  the  most  important 
of  the  differential  equations  of  physics,  namely,  the  differential 
equation  of  wave  motion. 

Let  the  expression  x  —  vt  be  represented  by  the  single  letter  z. 
That  is 

z  =  X  —  vt  (4) 

so  that 

dz 


and 


Then  equation  (2)  becomes 

y  =  F(z)  ^  ^  (7) 

Let  ^  be  represented  by   F'(z),   and  let   ^  [  =  ^^^^'l     be 

represented  by    F"{z).    Then,  according  to  the  rule  for  differ- 
entiating a  function  of  a  function  (see  Art.  34),  we  have  (using 


the  above  values  of  —  and   —  ) : 
dx  dt  I 


dy  _dy     dz  _     ,.  .  .  . 

Fx-Tz'd-x-^^'^  (^> 


differentiating  again  we  have : 


S=«5^-l= '"•(•)  ») 


Similarly  we  have: 


S4M=--'(^)  ao) 


192  CALCULUS, 

and 

Therefore,  comparing  equations  (9)  and  (11),  we  have 

and  it  may  be  shown,  as  above,  that  equation  (3)  leads  to  this 
same  differential  equation. 

A  partial  differential  equation  may  be  recognized  as  one  which 
contains  more  than  one  independent  variable.  Therefore  it  is 
unnecessary  to  use  the  symbol  d  instead  of  the  symbol  d.  Here- 
after the  symbol   d  will  be  used  exclusively. 

General  solution  of  eqvxition  {12). — Equations  (2)  and  (3)  both 
satisfy  equation  (12),  and  therefore  equations  (2)  and  (3)  are  both 
solutions  of  (12).  Therefore,  according  to  the  principle  of  super- 
position as  explained  in  Art.  100,  the  sum  of  F{x  —  vt)  and 
fix  +  vt)   or 

y  =  F(x-vt)+f{x  +  vt)  (13) 

is  a  solution  of  equation  (12).     Indeed  this  is  the  general  solution 
of  (12)  because  it  contains  two  undetermined  fimctions.* 

PROBLEMS. 
l.Findthevaluesof  |,3,Jandgwhen 

y  =  A  sin  (x  —  vt); 
also  when 

y  =  A  sin  (x  -\-  vt). 

Show  that  both  of  these  functions  satisfy  equation  (12)  of  Art.  107. 

*  The  general  solution  of  an  ordinary  differential  equation  of  the  second 
order  contains  two  undetermined  constants,  and  the  general  solution  of  a 
partial  differential  equation  of  the  second  order  contains  two  undetermined 
functions. 


DIFFERENTIAL  EQUATION  OF  WAVE  MOTION.  193 

2.  Show  that 

A  sin  {x  —  yQ  +  A  sin  (x  +  vi)  =  2A  sin  x  cos  vi^ 

and  plot  the  curve    y  =  2A  sin  x  cos  vi    for  the  following  values 
of  vtj   namely, 

0,  ^,  2'  ^»  'T^  -4  »  -2^  -^    and   ^tt. 

iVote. — The  curves  thus  plotted  show  the  successive  configurations  of  a 
string  which  is  vibrating  in  what  is  called  a  simple  mode.  See  Franklin  and 
MacNutt,  Light  and  Sound,  page  243,  The  Macmillan  Co.,  New  York,  1909 

108.  Equation  of  motion  of  a  stretched  string.  —  When  a 
stretched  string  is  in  equilibrium  it  is  of  course  straight.  Let 
us  choose  this  equilibrium  position  of  the  string  as  the  a:-axis  of 
reference  as  shown  in  Fig.  90,  and  let  us  set  up  the  differential 


equation  which  expresses  the  mode  of  motion  of  the  string  while 
the  string  is  vibrating  or  while  a  bend  is  traveling  along  the  string 
as  a  wave.  We  will  assume  that  each  part  of  the  string  moves 
only  up  and  down  (at  right  angles  to  the  a;-axis  in  Fig.  90)  and  we 
will  assume  that  the  string  is  perfectly  flexible.*  Under  these 
conditions  the  a;-component  of  the  tension  of  the  string  is  always 
and  everywhere  equal  to  a  constant   T. 

Let  the  curve  ccc,  Fig.  90  be  the  configuration  of  the  string 
at  a  given  instant  t,  that  is,  ccc  is  what  a  photographer  would 
call  a  snapshot  of  the  moving  string.    The  shape  of  the  curve  ccc 

*  This  means  that  the  only  thing  that  keeps  the  string  straight  is  its  tension. 
14 


194  CALCULUS. 

defines  y  as  a.  function  of  x,  and  the  steepness  of  the  curve  at  a 

dv 
point  is  the  value  of   -^  at  that  point. 

Consider  the  very  short  portion  ab  of  the  string.  The  length 
of  this  portion  of  the  string  when  the  string  lies  along  the  rc-axis 
(in  equilibrium)  is  dx,  and  the  mass  of  the  portion  is  m  -  dx 
pounds  or  grams  as  the  case  may  be,  where  m  is  the  mass  per  unit 
length  of  the  string.  An  enlarged  view  of  the  very  short  portion 
of  the  string  is  shown  in  Fig.  91.  The  adjacent  parts  of  the 
string  pull  on  the  portion  ab  in  the  direction  of  the  string  at  a 
and  at  6,  and  the  forces  R  and  R'  are  thus  exerted  on  the 
portion  ah.  The  a;-component  of  R  is  the  force  T  towards  the 
left,  and  the  a;-component  of  R'  is  an  equal  force  T  towards  the 
right.  Therefore  the  downward  force  D  on  the  end  a  of  the 
portion  ah  \s  D  =  T  tan  B,  and  the  upward  force  U  acting  on 
the  end  h  is  U  =  T  tan  6'.  Therefore  the  net  upward  force 
exerted  on  ah  is 

dF  =  T  tan  6'  -  T  tan  0  (1) 

dv 
But  tan  6  is  equal  to  the  value  of  -^  at  a,  and  tan  0'  is  equal 

dv 
to  the  value  of  -p  at  h.    Therefore  the  difference,  tan  0'—  tan  0, 

dv 
is  the  increase  of  -^  from  a  to  h,   and  this  increase  is  equal  to 

cPv  d^v 

-t4  •  dx.    This  is  evident  when  we  consider  that  the  value  of  -j^ 

dv 
means  the  rate  of  increase  of  -f-  with  respect  to    x.    Therefore, 

d^v 
substituting  j^  •  dx  for     tan  6'  —  tan  d     in  equation  (1),  we 

have: 

dF  =T^^'dx  (2) 

Now  according  to  Newton's  laws  of  motion  the  net  upward 
force   dF   acting  on  the  portion   ah   of  the  string  is  equal  to  the 


^<^- 


DIFFERENTIAL  EQUATION  OF  WAVE  MOTION.         195 


mass  m  '  dx  oi  the  portion  multiplied  by  the  upward  acceleration 
-z^    of  the  portion.     The 
in  equation  (2),  we  have: 


-z^    of  the  portion.     Therefore,  substituting    m  -^  '  dx    for    dF 


d^  d^ 

^  dt'       ^  dx' 
or 

dp-       m  dx^  ^^ 

This  differential  equation  is  here  derived  for  the  case  of  wave 
motion  on  a  string,  but  identically  the  same  form  of  equation 
applies  to  sound  waves  in  air,  to  electric  waves,  to  waves  of  light 
(which  indeed  are  electric  waves),  and  to  certain  kinds  of  water 
waves.*  Equation  (3)  is  therefore  very  important.  Its  general 
solution  as  found  in  Art.  107  is: 

y  =  F{x-vt)+f(x  +  vt)  (4) 

where  _ 

V  =  a/-  (5) 

The  term  F(x  —  vt)  represents  a  wave  (a  bend  of  any  shape) 
traveling  to  the  right  at  velocity  v^  and ,  the  term  f{x  +  vt) 
represents  a  wave  (a  bend  of  any  shape)  traveling  to  the  left  at 
velocity   v. 

109.  Idea  of  wave  motion  established  without  the  help  of 
equation  (3)  of  Art.  108. — It  is  evident  from  Art.  107  that  travelj 
purely  and  simply,  is  about  the  only  thing  that  is  estabUshed  by 
the  solution  of  equation  (3)  of  Art.  108.  Therefore  one  might 
expect  to  obtain  equations  (4)  and  (5)  without  the  help  of  equation 
(3)  by  introducing  the  idea  of  travel  at  the  beginning.  With  this 
end  in  view  let  us  consider  a  bend  of  any  shape  and  let  us  imagine 
that  this  bend  is  traveling  along  the  string  to  the  right  at  any 

*  We  are  here  considering  only  the  simple  case  in  which  there  are  no  appreci- 
able energy  losses  as  the  wave  travels  along. 


196  CALCULUS. 

velocity  v.  One  could  make  a  bend  of  definite  shape  travel 
along  a  stretched  string  by  threading  the  string  through  a  bent 
tube  and  moving  the  bent  tube  along,  and  this  state  of  affairs 
would  be  entirely  unchanged  if  we  imagine  the  tube  to  he  stationary 
and  the  stretched  string  to  he  drawn  through  it  as  indicated  in  Fig.  92. 


Fig.  92. 

It  is  assumed  that  the  string  slides  through  the  tube  without 
friction.  At  any  point  p  the  string  pushes  against  the  side  d 
of  the  tube  because  of  its  tension,  and  the  string  pushes  outwards 
against  the  side  c  of  the  tube  because  of  centrifugal  action. 
Let  r  be  the  radius  of  curvature  of  the  tube  at  p.  Then  the 
particles  of  the  string  near  p  may  be  considered  to  travel  along 
a  circular  path  of  radius  r.    Therefore,  according  to  Art.  50,  the 

radial*  acceleration  of  a  particle  of  the  string  at  p  is  - .    Consider 

r 

an  infinitely  short  piece  of  the  string  whose  length  is    ds    and 

whose  mass  is    m  •  ds.    Then    m  '  ds  X  -     is  the  radial  force 

r 

which  must  act  upon  the  short  piece  of  string  to  produce  the 

specified  acceleration  — .    If  the  string  has  no  tension,  then  all  of 

this  radial  force  is  exerted  by  the  side   c   of  the  tube. 

If  the  string  is  under  tension    T,    then  the  tension  produces  a 

T 
radial  force  equal  to    -  •  ds    on  the  portion    ds    of  the  string, 

according  to  Art.  51;  and  if  the  string  is  not  moving  this  force  is 
exerted  against  the  side  d  of  the  tube. 

*  In  the  direction  of  r  and  towards  the  center  of  the  osculating  circle. 


DIFFERENTIAL  EQUATION  OF  WAVE  MOTION.         197 
If  the  string  is  moving  at  a  velocity  which  satisfies  the  equation: 


or 


V'       T 
m  -  =  - 
r       r 


(1) 


then  the  radial  force  due  to  the  tension  of  the  string  is  just  sufficient 
to  produce  the  necessary  radial  acceleration,  and  no  force  at  all 
is  exerted  on  the  string  by  the  guiding  tube.  Under  these  con- 
ditions the  guiding  tube  might  be  removed  and  the  bend  would 
stand  in  a  fixed  position  on  the  traveling  string;  or  if  the  string 
were  standing  still  the  bend  would  travel  along  the  string  at 
velocity  v. 

In  this  discussion  the  bend  can  be  of  any  shape  and  it  can 
travel  at  velocity  v  in  either  direction. 

110.  The  vibration  of  a  plucked  string.* — A  string  is  pulled 
to  one  side  as  shown  in  Fig.  93  and  released.    The  string  is  thus 


Fig.  93. 

*  The  vibration  of  a  string  which  is  struck  with  a  hammer  as  in  a  piano  is 
easy  to  formulate.  The  vibration  of  a  string  which  is  set  vibrating  by  a  bow 
as  in  a  violin  is  somewhat  more  difficult  to  formulate. 

A  good  discussion  of  this  subject  is  given  in  Byerly's  Fourier's  Series  and 
Spherical  Harmonics,  pages  30-145,  Ginn  and  Co.,  Boston,  1893. 

A  discussion  of  the  motion  of  piano  strings  and  of  violin  strings  is  given 
in  Appendices  V  and  VI  of  Helmholtz's  Tonempfindungen.  English  trans- 
lation by  Alexander  J.  Ellis  is  entitled  Sensations  of  tone.  Published  by 
Longmans,  Green  and  Co.,  1885. 


198  CALCULUS. 

set  vibrating.  The  vibrating  string  has  at  each  instant  a  definite 
shape,  that  is,  it  forms  a  definite  curve.  It  is  desired  to  find  an 
equation  expressing  y  in  terms  of  x  and  elapsed  time  t,  which 
equation  will  be  at  each  instant  the  equation  of  the  curve  formed 
by  the  moving  string. 

There  are  two  conditions  which  must  be  satisfied  by  the  ex- 
pression for  ?/,  namely:  (a)  y  must  be  zero  at  all  times  when 
x  =  0  and  when  x  =  tt,  and  (6)  at  the  instant  t  =  0  the  ex- 
pression for  y  must  be  the  equation  of  the  curve  formed  by  the 
plucked  string  at  the  moment  of  release. 

Also,  of  course,  the  expression  for  y  must  satisfy  the  differential 
equation  (3)  of  Art.  108,  namely, 

(Py^T^ 

dt^      rndx"  ^^ 

Now  it  can  be  shown  (see  problem  2  on  page  193)  that  the  following 
expressions  satisfy  equation  (1) 

y  =  A  sin  nx  sin  nvt  (2) 

2/  =  A  sin  nx  cos  nvt  (3) 

y  =  A  cos  nx  sin  nvt  (4) 

y  =  A  cos  nx  cos  nvt  (5) 

where  A  and  n  have  any  values  whatever.  But  only  (2)  and 
(3)  give  ?/  =  0  when  a;  =  0,  and  n  must  he  an  integer  to  give 
2/  =  0  when  x  =  w.  But  (2)  cannot  be  used  because  it  gives 
2/  =  0  everywhere  when  ^  =  0.  Therefore  our  problem  is  to 
take 

2/  =  A  sin  nx  cos  nvt  (3) 

which  satisfies  condition  (a),  and  add  a  large  number  of  such  solu- 
tions together  (using  different  values  of  A  and  n  in  each)  to 
get  an  expression  for  y  which  satisfies  condition  (6),  above.  The 
possibility  of  doing  this  was  discovered  by  Fourier*  and  is  embodied 
in  what  is  called  Fourier's  theorem. 

*  Fourier's  original  discussion  is  very  simple  and  interesting.  See  Fourier's 
Theory  of  Heal,  translated  by  Alexander  Freeman,  Cambridge,  1878. 


DIFFERENTIAL  EQUATION  OF  WAVE  MOTION.         199 

111.  Fourier's  theorem. — ^Let  y  be  any  given  function  of  x. 
Consider  a  certain  portion  AB  of  the  curve  which  represents  y 
as  shown  in  Fig.  94,  and  let  the  distance    AB    be    27r    (this  is 

AB 
equivalent  to  choosing  -^r—  as  our  unit  of  length).     Then,  ac- 
cording to  Fourier  we  may  express  the  function    y    within  the 
region  AB  as  follows: 


y  =  Ao  -{-  Ai  sin  X  +  Az  sin  2x  +  ^3  sin  3x  + 
+  Bi  cos  X  +  B2  cos  2x  +  B3  cos  Sx 


::::) 


(1) 


where  the  coefficients   Ao,  Ai,  A2,  •  •  •    and    Bi,  B2,  B3,  •  •  •    are 
constants  whose  values  are : 


1        /»a;=2rr 

Ao  =  ^  J        ydx  (2) 

-j        r*x=2ir 

An='  ~  \         y  minx  '  dx  (3) 

'^  Jx=0 
-t       nx=Z7r 

—  I         y  cosnx  '  dx  (4) 

'^  Jx=Q 


=277 

B.      I 

'"    lx=Q 


A  complete  proof  of  Fourier's  theorem  would  involve  a  proof 
that  the  right-hand  member  of  equation   (1)   is  a  convergent 


\^^' 


series.*    This,  however,  will  here  be  taken  for  granted.    The 
proof  then  reduces  to  a  derivation  of  equations  (2),  (3)  and  (4). 

*  This  matter  is  discussed  in  Chapter  III  of  Byerly's  Fourier's  Series  and 
Spherical  Harmonics,  Ginn  and  Co.,  Boston,  1893. 


200  CALCULUS. 

In  this  proof  a  consideration  of  average  values  is  most  important, 

and  the  student  should  therefore  look  over  Art.  73  again. 

Another  matter  of  import- 
ance is  to  understand  clearly 
'V^A^uin  fix  ^^  ^YiQ  meaning  of  the  equation 
y  =  sin  nx  (and  y  =  cos  nx) 
when  n  is  an  integer.  Now 
y  —  sin  nx  is  the  equation 
of  a  sine  curve.  If  n  =  1 
this  sine  curve  makes  a  posi- 
tive "arch"  between  x  =  0 

and  a;  =  TT,  and  a  negative  "arch'*  between  x  =  t  and  x  =  27r. 

In  general  the  sine  curve     y  =  sin  nx     makes  a  positive  arch 

between  x  =  0  and  x  =  -,  and  a  negative  arch  between  x  =  - 

n'  ^  n 

and  X  =  —   as  shown  in  Fig.  95.    Therefore  the  sine  curve 

n  ^ 

y  =  sinnx  makes  n  pairs  of  positive  and  negative  arches  between 
X  =  0  and  x  =  2t, 

The  derivation  of  equations  (2),  (3)  and  (4)  depends  upon  the 
following  propositions,   n  and  m  being  integers. 

Proposition  I. — The  average  value  of  sin  nx  (or  of  cos  nx)  between 
X  —  0  and  x  =  ^tt  is  zero.  This  is  evident  when  we  consider 
that  sin  nx  (or  cos  nx)  passes  through  exactly  similar  sets  of 
positive  and  negative  values  which  exactly  offset  each  other 
between  a;  =  0  and  x  =  27r. 

Proposition  n. — The  average  value  of  sin^  nx  (or  of  cos^  nx) 
between    x  =  0    and   x  =  2x   is  equal  to  \.    This  may  be  shown 

-I  /*Z=ilt 

by  finding  the  value  of  ^   I  m^nx-dx  (or  of  the  correspond- 

ing  expression  for  cos^  nx)  as  explained  in  Art.  73. 

Proposition  m. — The  average  value  of  sin  nx  sin  mx  between 
X  =  0   and  x  =  ^ir  is  zero  when   n   and  m  are  different  integers. 


DIFFERENTIAL  EQUATION  OF  WAVE  MOTION.         201 
This  may  be  shown  by  finding  the  value  of  the  expression 

'oZ  I  sin  nx  sin  mx  •  dXy 

which  may  be  easily  done  with  the  help  of  form  24  in  the  table 
of  integrals. 

Proposition  IV. — The  average  value  of  sin  nx  cos  mx  between 
X  =  0  and  x  =  2t  is  zero  whether  the  integers  n  and  m  he  the 
same  or  not.  This  may  be  shown  with  the  help  of  form  25  in  the 
table  of  integrals. 

Proposition  V. — The  average  value  of  cos  nx  cos  mx  from  x  =  0 
to  X  =  2t  is  zero  when  m  and  n  are  different  integers.  This  can 
be  shown  with  the  help  of  form  26  in  the  table  of  integrals. 

Derivation  of  equation  (2). — Consider  the  average  value  of  each 
member  of  equation  (1)  between  x  —  0  and  x  =  27r.  The 
average  value  of  the  first  member  is,  by  definition,  equal  to 

—  I         y-dx,   and  the  average  value  of  the  second  member  is  Ao 

according  to  proposition  /  above.     Therefore  ^  I         y'dx  =  Ao 

which  is  equation  (2). 

Derivation  of  equation  (3). — Multiply  both  members  of  equa- 
tion (1)  by  sin  nx  and  we  have: 

y  sin  nx  =  Ao  sin  nx  +  Ai  sin  nx  sin  x  -{-  •  •  •  -{-An  sin^  nx 

+  •  •  •  +  -Bi  sin  na;  cos  X  +  •  •  • . 

Consider  the  average  value  of  each  member  of  this  equation 
between    a;  =  0    and    x  =  27r.    The  average  value  of  the  first 

member  is  ;r-  (         y  sin  nx-dx.    The  average  value  of  every  term 

^T^  Jx=0 

of  the  second  member  is  zero  according  to  the  above  propositions 
except  the  term    An  sin^  nx,    and  the  average  value  of  this  term 


202 


CALCULUS. 


is  }An  according  to  proposition  IL    Therefore 
^1         y  sinnx  '  dx  =  iAn 

which  is  equation  (3) 

Derivation  of  equation  (4). — ^Multiply  both  members  of  equa- 
tion (1)  by  cos  nx,  consider  the  average  value  of  each  member 
of  the  resulting  equation  between  a;  =  0  and  x  =  27r,  and 
equation  (4)  is  obtained. 

Simplification  of  equations  (3)  and  (4)  due  to  symmetry  of  given 
curve. — Let  ccc,  Fig.  96a,  be  the  curve  which  is  to  be  expressed 
by  equation  (1).    It  is  permissible  to  take  the  base  of  ccc  as  tt. 


Fig.  96a. 

Under  these  conditions  there  are  two  important  methods  for 
choosing  the  second  half  of  the  complete  curve,  namely,  the 
portion  between  tt  and  27r,   as  follows: 

Curves  having  sine  terms  only. — The  second  half  of  the  complete 
curve  may  be  chosen  like  ccc  turned  upside  down  and  turned 
end  for  end,  as  shown  by  the  dotted  curve  c'c'  in  Fig.  96a.  In 
this  case  y  =  —  y'  as  shown  in  Fig.  96a.  Now  sin  nx  has  equal 
and  opposite  values  at  p  and  at  p',  therefore  y  sin  nx  •  dx  has  the 
same  value  and  sign  at  p   and  at  p',   and  therefore 


ry  %\Ti.  nx  •  dx  =  2  \     y  ^\ 


sin  nx  •  dx 


DIFFERENTIAL  EQUATION  OF  WAVE  MOTION.         203 
Consequently  equation  (3)  becomes: 

An  =  -  I     y  sinnx  '  dx  (5) 

TT  Jo 

Furthermore   cos  nx  has  the  same  value  and  the  same  sign  at   p 
and  at    p',    therefore   y  cos  nx  •  dx    has  the  same  value  but  has 

/»27r 

opposite  signs  at  p  and  at  p\  and  therefore     I     y  cos  nx  •  dx   is 

zero.     Consequently  all  of  the  cosine  terms  drop  out  of  equation 
(1).     Furthermore  it  is  evident  from  the  symmetry  of  the  com- 

plete  curve    cccc^c'    in  Fig.  96a  that      1      y-dx  =  0    so  that 

Jo 
Ao  =  0.    Therefore  the  complete  curve  in  Fig.  96a  is  given  by: 

y  =  Ai  sin  X  +  Az  sin  2x  +  A3  sin  3x  +  •  •  •  (6) 

Curves  having  cosine  terms  only. — The  second  half  of  the  complete 
curve  may  be  chosen  like  ccc  turned  end  for  end  but  not  turned 
upside  down  as  shown  by  the  dotted  curve    c'c'c'    in  Fig.  966. 


y'tixia 


K 


x-axis 


^    0' 


1 


—  X -ii  !  K (x) 

^ — . — ~-;r-^ ^ 

Fig.  966. 


Under  these  conditions  it  may  be  shown,  by  an  argument  some- 
what similar  to  the  above,  that  equations  (1)  and  (4),  respectively, 
become 

y  =  Ao  +  Bi  cos  X  -\-  B2  cos  2x  +  B3  cos  3a;  +  •  •  •         (7) 

and 

2   f' 

Bn  =  ~  \     y  COS  nx  '  dx  (8) 


204 


CALCULUS. 


112.  The  vibration  of  a  plucked  string  completely  formulated. — 
In  Art.  110  the  problem  of  the  plucked  string  was  reduced  to  the 
problem  of  adding  together  a  number  of  terms  like  A  sin  nx  cos  nvt 
so  as  to  get  an  expression  for  y  which  is  the  equation  of  the  curve 
formed  by  the  string  when  t  =  0.  This  is  evidently  the  same 
thing  as  expanding  the  given  function  y  '\n  o.  series  of  sines  as 
explained  in  Art.  HI.  That  is,  we  must  find  the  coefficients  in 
the  series: 

2/  =  ill  sin  a;  +  Ai  sin  2x  +  ^la  sin  3a;  +  •  •  •  (1) 


and  the  general  expression  for  the  coefficient  An  is: 
An  =  -  \     2/  sin  nx  •  dx 


(2) 


according  to  equation  (5)  of  Art.  Ill,  the  use  of  which  means 
that  we  have  assumed  the  length  of  our  string  as  tt  units  and 

middle  of  string 
-string 


Fig.  97o. 

have  completed  the  curve  as  indicated  by  the  heavy  dotted  line 
in  Fig.  97a. 

Now  the  equation  of  the  curve  formed  by  the  string  at  the 
beginning   {t  =  0),   as  shown  in  Fig.  97a,  is: 


y 


2c 


from    X  =  0    to    ^  =  -^ 


2c 


y  =  2c X    from    x  =  ^    to    x  = 

IT  Z 


(3) 
(4) 


DIFFERENTIAL  EQUATION  OF  WAVE  MOTION.         205 
Therefore  the  integral  (2)  must  be  evaluated  in  two  parts,  namely: 

An  =  -  t       —  xsmnx  '  ax  +  -  I     [2c x  ]  smnx-dx     (5) 

which  gives: 

.         Scl    .    nx 

ir^w-         2 

Therefore  the  equation  of  the  curve  formed  by  the  string  at  the 
beginning  is: 

y  =  -^  (sin  x  —  ^  sin  3x  +  ^V  sin  5a;  —  •  •  • )  (6) 

and  the  equation  of  the  curve  formed  by  the  string  at  an  instant 
t  seconds  after  the  beginning  is: 

y  =  —^  (sin  X  cos  vt  —  ^  sin  Sx  cos  3vt  ■}-  •  •  • )  (7) 

Description  of  the  motion  of  a  plucked  string. — It  would  seem 
from  the  elaborate  difficulties  of  the  above  problem  [and  perhaps 

>qTTTTTTTpr 
TTTTTlXTTrt\^  ^ 

"T^iXnTTTTTTTTTTpf 

Fig.  976. 


206 


CALCULUS. 


the  complexity  of  the  result  as  given  in  equation  (6)  would  suggest] 
that  the  motion  of  a  plucked  string  is  very  complicated.  As  a 
matter  of  fact,  however,  the  motion  of  a  plucked  string  is  extremely 
simple  and  easy  to  describe.  Thus  Fig.  976  shows  six  successive 
snapshots  of  a  string  which  has  been  plucked  in  the  middle  and 
released.  The  moving  part  of  the  string  is  straight  and  parallel  to 
the  equilibrium  position  of  the  string  and  the  motion  is  indicated 
by  the  short  arrows. 

PROBLEMS. 

The  most  important  practical  problem  involving  Fourier's  theorem  is, 
perhaps,  the  finding  of  the  coefficients  in  equation  (1)  of  Art.  Ill  when  the 
function  y  is  given  not  as  an  algebraic  function  of  x  but  as  an  experimentally 
determined  curve. 

A  very  good  set  of  directions  for  making  the  necessary  calculations  is 
given  in  Bedell  and  Pierce's  Direct  and  Alternating  Current  Manual,  2nd  edition, 
pages  331-338,  D.  Van  Nostrand,  New  York,  1913;  and  a  discussion  of  the 
origin  and  proof  of  the  method  is  given  on  pages  339-344. 

The  harmonic  analyzer  is  a  machine  for  finding  the  coefficients  in  equation 
(1)  of  Art.  Ill  when  the  function  y  is  given  as  an  experimentally  determined 
curve.  The  eariiest  harmonic  analyzer  is  that  of  Lord  Kelvin.  A  brief 
description  of  this  machine  is  given  in  Franklin's  Electric  Waves,  pages  340- 
342,  The  Macmillan  Co.,  New  York,  1909.  See  also  section  37  of  the  article 
on  Tides  in  the  ninth  edition  of  the  Encyclopedia  Britannica.  See  also  articles 
by  James  Thomson  and  by  Sir  Wm.  Thomson  (Lord  Kelvin)  in  Proceedings  of 
the  Royal  Society,  Vol.  XXIV,  1876,  page  262  and  pages  269  and  271.  The 
harmonic  analyzer  of  G.  U.  Yule  is  described  in  an  article  by  J.  N.  Le  Conte, 
Physical  Review,  Vol.  VII,  pages  27-34,  1898.  An  especially  interesting  de- 
scription of  a  harmonic  analyzer  is  given  on  pages  68-74  of  A.  A.  Michelson's 
Light  waves  and  their  uses,  University  of  Chicago  Press,  Chicago,  1903. 

1.  Determine  the  coefficients  in  Fourier's  series  to  give  the 
curve  cc  which  is  shown  in  Fig.  p\. 


Fig.  pi. 


DIFFERENTIAL  EQUATION  OF  WAVE  MOTION.         207 

Ans:  y  = (sin  x  +  J  sin  2a;  +  J  sin  3a;  +  •  •  • ) 

TT 

2.  Determine  the  coefficients  in  Fourier's  series  to  give  the 
curve  cc  which  is  shown  in  Fig.  p2. 

Ans;  y  =  -^  (sin  a;  +  J  sin  3a;  +  J  sin  5a;  +  •  •  •) 

TT 


y-axis 

c 

T 

r 

1 

c     \a 

c 

1 

1 
* 

!       x-ojcw 

1 

1 

1 

c     \a 

c 

' 

i      c 

h 


K -^ -J(— ^- — ^ 


Fig.  p2. 


3.  Determine  the  coefficients  in  Fourier's  series  to  give  the 
curve   cc  which  is  shown  in  Fig.  p3. 

Ans:  y  =  —2  (cos  a;  —  |  cos  3a;  +  i  cos  5a;  —  |  cos  7a;  +  •  •  •) 


y-axis 
e 

c 

n            r  — 

f 
X 

c                       c 
x-curis 

1 

I 
1 

1 

L                          J 

c 

1 
la 

c 

1 

i 7( *-^ 

, 

Fig.  p3. 


4.  Determine  the  coefficients  in  Fourier's  series  to  give  the 
curve  cc  which  is  shown  in  Fig.  p4. 

Ans:  y  =  —  (sin  a;  —  J  sin  3a;  +  -jV  sin  5a;  —  :jV  sin  7a;  +  •  •  •) 


208 


CALCULUS. 


5.  Determine  the  coefficients  in  Fourier's  series  to  give  the 
curve  cc  which  is  shown  in  Fig.  p5. 


y-€ixi8 


Sa 


Ads:  y  =  -^  (cos  a;  +  i  cos  3a;  +  ^  cos  5a;  +  •  •  • ) 

y-axis 

./ 

"  ^^          »-axw         ^     1      \. 

■■--.--'• 

^^\i>^       1       "^^-^^- 

< — n ^—-n i 

Fig.  p5. 

6.  Determine  the  coefficients  in  Fourier's  series  to  give  the 
curve  cc  which  is  shown  in  Fig.  p6. 

Ans:  y  =  —  (sin  a;  —  J  sin  2a;  +  J  sin  3a;  —  i  sin  4a;  +  •  •  • ) 


t 


DIFFERENTIAL  EQUATION  OF  WAVE  MOTION.         209 

7.  Derive  the  answer  to  problem  4  by  integrating  the  answer 
to  problem  3. 

Note. — Any  Fourier  series  gives  a  convergent  series  when  integrated  term 
by  term. 

8.  Plot  the  curve  which  represents  the  integral  of  the  curve 
given  in  Fig.  p2,  the  constant  of  integration  being  such  as  to 
make  the  integral  curve  pass  through  the  origin;  and  determine 
the  coefficients  in  a  Fourier's  series  to  give  the  integral  curve. 

Note. — The  constant  of  integration  is  zero  in  problem  7  and  therefore 
easily  overlooked. 

9.  Derive  the  answer  to  problem  3  from  the  answer  to  problem  2 
by  shifting  the  origin  ^  to  the  right. 

10.  Differentiate  the  answer  to  problem  4  and  interpret  the 
result. 

Note- — A  Fourier  series  gives  a  divergent  series  by  differentiation  unless 
the  function  ij  is  continuous.  Thus  y  in  Fig.  p4  is  continuous,  whereas  y 
in  Figs,  pi,  p2,  p3  and  p6  is  discontinuous. 


15 


CHAPTER  IX 

VECTOR  ANALYSIS. 

SCALAR  AND  VECTOR  FIELDS. 

113.  Space  analysis. — A  system  of  space  analysis,  commonly 
called  vector  analysis,  is  developed  in  this  chapter,  and  it  is  ex- 
tremely useful  and  important  in  every  branch  of  physics  where 
variations  in  space  are  involved.  Thus  the  theory  of  electricity 
and  magnetism  (aside  from  the  electron  theory)  is  a  simple  appli- 
cation of  vector  analysis.  Also  the  theory  of  heat  flow,  the 
theory  of  fluid  motion,  and  a  great  part  of  the  theory  of  elasticity 
and  wave  motion  are  applications  of  vector  analysis.  It  is  almost 
impossible  for  a  student  to  make  a  beginning  in  any  branch  of 
theoretical  physics  without  some  understanding  of  vector  analysis. 

Vector  analysis  as  outlined  in  this  chapter  is  a  very  different 
thing  from  the  familiar  use  of  vectors  and  of  complex  quantity 
in  the  theory  of  alternating  currents.  In  the  one  case  we  have  a 
comprehensive  system  of  space  analysis,  and  in  the  other  case 
we  have  a  narrow  scheme  for  representing  the  solution  of  a  linear 
partial  differential  equation,  as  explained  in  chapter  VII.  Indeed 
the  kind  of  "vector  analysis''  which  is  used  in  the  theory  of 
alternating  currents  cannot  possibly  be  extended  to  three  dimen- 
sions as  a  consistent  system  of  space  analysis. 

Vector  analysis  originated  in  Sir  William  Hamilton's  theory  of 
quaternions.*  The  theory  of  quaternions,  however,  contains 
much  that  is  of  doubtful  value  in  theoretical  physics.  Indeed 
Maxwell  in  his  great  treatise  on  Electricity  and  Magnetism  (Oxford, 
1873)  used  only  a  few  of  Hamilton's  ideas.     The  study  of  Max- 

*  See  Sir  William  Rowan  Hamilton's  Lectures  on  Quaternions,  and  Elements 
of  Quaternions.  These  books  are  now  out  of  print  but  they  may  be  found 
in  most  good  libraries.  An  Elementary  Treatise  on  OuMernions  by  P.  G.  Tait, 
second  edition,  Oxford,  1873,  is  the  standard  treatise  on  the  subject. 

210 


VECTOR  ANALYSIS.  211 

well's  treatise  has  kept  the  subject  of  space  analysis  before  the 
serious  student  of  physics  for  more  than  a  generation,  and  two 
significant  attempts  have  been  made  to  formulate  the  more  useful 
of  Hamilton's  ideas  in  what  is  now  called  vector  analysis.  The 
first  of  these  was  that  of  Willard  Gibbs,  whose  view  of  vector 
analysis  was  outlined  in  a  very  condensed  form  in  a  pamphlet 
printed  (for  private  circulation,  only)  in  New  Haven  in  1883.* 
The  second  attempt  to  formulate  the  more  useful  of  Hamilton's 
ideas  was  that  of  Oliver  Heaviside.f 

The  first  attempt  to  place  vector  analysis  before  the  student  of 
physics  in  simple  form  was  that  of  E.  L.  Nichols  and  W.  S.  Frank- 
lin.J  The  best  discussion  of  vector  analysis  for  the  student  is 
that  of  Abraham  and  Foppl  in  their  Theorie  der  Electridtdt,  Vol.  I, 
pages  1-125,  Leipsig,  1907. 

114.  Scalar  and  vector  quantities. — A  scalar  quantity  is  a 
quantity  which  has  magnitude  only.  Thus  every  one  recog- 
nizes at  once  that  to  specify  10  cubic  yards  of  sand,  25  pounds  of 
sugar,  5  hours  of  time  is  in  each  case  to  make  a  complete  speci- 
fication. Such  quantities  as  volume,  mass,  time  and  energy  are 
scalar  quantities. 

A  vector  quantity  is  a  quantity  which  has  both  magnitude  and 
direction,  and  to  completely  specify  a  vector  quantity  one  must 
give  both  its  magnitude  and  its  direction.  This  necessity  of 
specifying  both  the  magnitude  and  the  direction  of  a  vector  is 
especially  evident  when  one  is  concerned  with  the  relationship  of 
two  or  more  vectors.  Thus  if  a  man  travels  a  stretch  of  10  miles 
and  then  a  stretch  of  5  miles  more,  he  is  by  no  means  necessarily 
15  miles  from  home.     If  T)ne  man  pulls  on  a  post  with  a  force  of 

*  Professor  Gibbs'  point  of  view  is  set  forth  in  Vector  Analysis  by  E.  B. 
"V^ilson,  New  Haven,  Yale  University  Press,  1901. 

t  See  Heaviside's  Electromagnetic  Theory^  Vol.  I,  pages  132-305,  The 
Electrician  Publishing  Co.,  London,  1893.  This  discussion  of  Heaviside'a 
is  unusually  interesting,  but  we  cannot  agree  with  Heaviside  in  his  statement 
that  vector  analysis  is  independent  of  the  quaternion. 

X  See  Nichols  and  FrankUn's  Elements  of  Physics,  Vol.  II,  first  edition,  The 
Macmillan  Co.,  New  York,  1895. 


212  CALCULUS. 

200  units  and  another  man  with  a  force  of  100  units,  the  total 
force  acting  on  the  post  is  by  no  means  necessarily  equal  to  300 
units.  A  New  Yorker  traveling  steadily  at  a  speed  of  60  miles 
an  hour  would  by  no  means  necessarily  reach  Boston  in  5  hours, 
because  he  might  be  traveling  in  some  other  direction. 

Many  cases  arise  in  physics  where  it  is  necessary  to  consider 
the  single  force  which  is  equivalent  to  the  combined  action  of 
several  given  forces;  where  it  is  necessary  to  consider  the  actual 
velocity  of  a  body  due  to  the  combined  action  of  several  causes, 
each  of  which  alone  would  produce  a  certain  amount  of  velocity 
in  a  given  direction;  and  so  on.  The  single  force  or  single  velocity 
is  in  each  case  called  the  vector  sum  or  the  resultant  of  the  given 
forces  or  given  velocities. 

The  addition  of  vector  quantities  is  exemplified  by  the  addition 
of  forces  as  follows:  Two  given  forces  are  represented  by  the 
lines  a  and  6,  in  Fig.  98,  and  the  sum  or  resultant  of  the  forces 
is  represented  by  the  diagonal  r  of  the  parallelogram  constructed 
on  a  and  h  as  sides.     It  is  evident  that  the  geometrical  relation 


between  a,  b  and  r  is  completely  shown  by  the  triangle  in  Fig. 
99,  in  which  the  line  which  represents  force  h  is  drawn  from  the 
extremity  of  the  line  which  represents  force  a,  and  r  is  the 
closing  line  of  the  triangle. 

The  geometrical  construction  of  Fig.  99  gives  a  method  of 
adding  any  number  of  forces  with  the  least  amount  of  compli- 
cation, as  shown  in  Fig.  100.  Thus  to  add  four  given  forces 
a,  h,  c  and  d:  draw  a  Hne  representing  force  a;  from  the  end  of 
this  line  draw  a  line  representing  force    b;    from  the  end  of  this 


VECTOR  ANALYSIS.  213 

line  draw  a  line  representing  force  c;  and  so  on.  Then  the  closing 
line  r  of  the  polygon  abed  represents  the  sum  of  the  forces  in 
magnitude  and  in  direction. 

To  represent  a  vector  in  an  alge- 
braic discussion  it  is  usually  most 
convenient  to  represent  the  vec- 
tor in  terms  of  its  components  in 
the  directions  of  three  chosen 
rectangular  axes  of  reference. 
Thus  if  X,  F,  Z  are  the  compo- 
nents of  a  given  force  F  (or  any  p.  .^ 
vector  whatever)  then  the  force  may 
be  represented  as 

F  =  X+Y  +  Z. 

In  this  expression  vector  addition  is  of  course  understood 
because  X,  Y  and  Z  are  vectors. 

■    115.  Vector  products.     Case  I.    Parallel  vectors. — The  product 

of  two  parallel  vectors  is  a  scalar.    This  fact  is  exemphfied  in 

every  branch  of  physics.     Thus  to  multiply  a  force  by  the  distance 

a  body  has  moved  in  the  direction  of  the  force  gives  the  work  done 

by  the  force,  and  work  is  a  scalar  quantity.     To  multiply  a  force 

by  the  velocity  with  which  a  body  moves  in  the  direction  of  the 

force  gives  the  power  developed  by  the  force,  and  power  is  a 

scalar  quantity.    The  square  of  the  velocity  of  a  body  determines 

its  kinetic  energy,  and  energy  is  a  scalar  quantity.    A  plane  area 

is  a  vector  quantity  and  its  vector  direction  is  the  direction  of  the 

normal  to  the  area.    To  multiply  the  area  of  the  base  of  a  prism 

by  the  altitude  of  a  prism  gives  the  volume  of  the  prism,  and 

volume  is  a  scalar  quantity. 

The  quotient  of  two  parallel  vectors  is  a  scalar  quantity.    Thus 

let   F  be  the  force  exerted  by  a  fluid  on  a  flat  surface  of  area   a. 

F 
Then  the  quotient   -  is  the  hydrostatic  pressure  of  the  fluid,  and 
a 

hydrostatic  pressure  is  a  scalar  quantity. 


214  CALCULUS. 

Case  II.  Orthogonal  vectors. — Vectors  which  are  at  right  angles 
to  each  other  are  said  to  be  orthogonal.  The  product  of  two 
orthogonal  vectors  is  a  third  vector  at  right  angles  to  both  of  the 
given  vectors.  Thus  to  multiply  the  length  by  the  breadth  of  a 
rectangle  gives  the  area  of  the  rectangle,  and  area  is  a  vector  as 
above  explained.  To  multiply  a  force  F  by  the  perpendicular 
distance  I  from  the  Une  of  action  of  the  force  to  a  chosen  axis 
gives  the  torque  action  IF  of  the  force  about  the  axis,  and  torque 
action  is  a  vector  quantity. 

The  quotient  of  two  orthogonal  vectors  is  a  third  vector  at  right 
angles  to  both  of  the  given  vectors.  Thus  to  divide  the  area  of  a 
rectangle  by  the  length  of  one  side  gives  the  length  of  the  other 
side. 

Case  III.  Oblique  vectors. — The  product  of  two  oblique  vectors 
consists  of  two  parts;  one  part  is  a  scalar  and  the  other  part  is 
a  vector.  This  proposition  is  a  necessary  result  of  the  above 
statements  concerning  the  product  of  parallel  vectors  and  the 
product  of  orthogonal  vectors.  Thus  Fig.  101  shows  two  obUque 
vectors    U    and    V.    The  vector    V    can  be  resolved  into  the 

">^  components  7'  and  V"  parallel  to  U 

vy\    j  and  perpendicular  to   U    respectively, 

y^  ^\  1      ^  and  then     7'  +  V"     can  be  substi- 

v^     ^^^  ^1        tuted  for    V    in  the  product  UV,  giv- 

— ^— '        ing    UV  +  VV".     But    UV   is  the 

pj     jQj  product  of  two  parallel  vectors  and  it 

is  a  scalar,  and  UV"  is  the  product 
of  two  orthogonal  vectors  and  it  is  a  vector. 

From  Fig.  101  it  is  evident  that  V  =  V  cos  6  and  that 
7"  =  7  sin  6.  Therefore  we  have  the  following  important 
propositions: 

(a)  The  scalar  part  of  the  product  of  two  oblique  vectors  is 
equal  to  the  product  of  the  numerical  values  of  the  respective 
vectors  and  the  cosine  of  the  angle  between  them. 

(6)  The  numerical  value  of  the  vector  part  of  the  product  of 
two  oblique  vectors  is  equal  to  the  product  of  the  numerical  values 


VECTOR  ANALYSIS. 


215 


of  the  respective  vectors  and  the  sine  of  the  angle  between  them; 
and  of  course  the  direction  of  the  vector  part  of  the  product  is 
at  right  angles  to  the  plane  of  the  given  oblique  vectors. 

There  are  but  few  cases  in  physics  where  the  scalar  part  and  the 
vector  part  of  the  product  of  two  oblique  vectors  are  both  important, 
although  there  are  many  cases  where  the  scalar  part  of  a  vector 
product  is  important,  and  many  other  cases  where  the  vector  part 
of  a  vector  product  is  important.  Thus  a  force  F  acts  on  a  car 
as  shown  in  Fig.  102.     Imagine  the  car  to  move  a  distance   I   in 


Fig.  102. 


the  direction  of  the  track,  then  the  work  done  by  the  force  is  the 
scalar  part  of  the  product  of  the  two  obhque  vectors  F  and  l^ 
or:  the  work  done  is  equal  to  numerical  value  of  I  X  numerical 
value  of  F  X  cos  6.  The  vector  part  of  the  product  IF  in  this 
case  has  no  very  important  meaning. 

A  force  F  acts  on  a  crank  as  shown  in  Fig.  103.  The  torque 
action  of  the  force  about  the  axis  of  the  crankshaft  is  the  vector 
part  of  the  product  of  the  two  oblique  vectors  F  and  I;  the 
numerical  value  of  the  torque  is  equal  to  numerical  value  of  I  X 
numerical  value  of  F  X  sin  6,  and  the  direction  of  the  torque, 
considered  as  a  vector,  is  parallel  to  the  axis  of  the  shaft. 

Exact  direction  of  the  vector  part  of  a  vector  product. — ^The 
product  UV  in  Fig.  101  is  a  vector  at  right  angles  to  the  plane 
of  the  paper,  but  is  it  towards  the  reader  or  away  from  the  reader? 
Consistency*  requires  us  to  admit  that  if  the  product    UV"    is 

*  This  is  shown  by  the  following  discussion  of  Fig.  104  and  by  the  inter- 
pretation of  equation  (1).  Sir  William  R.  Hamilton  gives  a  very  full  discus- 
sion of  this  in  his  Lectures  and  in  his  Elements  of  Quaternions. 


216 


CALCULUS. 


towards  the  reader  in  Fig.  101,  then   V'U  must  be  away  from  the 

reader.    That  is,  we  must  admit  that    UV"  =  —  V"U    where 

U  and   V"  are  orthogonal  vectors. 
This  matter  is  exemphfied  as  follows:  A  force    F    acts  upon  a 

crank-arm  I  as  shown  in  Fig.  104.    The  vector  part  of  the  product 

IF  is  the  torque  action  of  the  force 
about  the  axis  0  (perpendicular  to  the 
plane  of  the  paper).  Now  the  torque 
action  of  F  in  Fig.  104  can  be  ex- 
pressed in  terms  of  the  components  of 
F,  namely,  X  and  F,  and  the  com- 
ponents of  I,  namely,  x  and  y. 
Thus  xF  is  a  counter-clockwise  torque 
about  0,  and  yX  is  a  clockwise  torque 


about  0  in  Fig.  104. 
as  positive,  we  have*. 


Therefore,  taking  counter-clockwise  torques 


Total  torque  action  of  F 
about  0  in  Fig.  104,  or  the 
vector  part  of  the  product  IF 


I  =a;y- 


yx 


(1) 


The  vector  direction  of  a  torque  may  be  thought  of  as  the  direc- 
tion (along  the  axis  of  the  torque)  in  which  a  right-handed  screw 
would  travel  if  turned  by  the  torque.  Adopting  this  convention 
and  choosing  the  positive  direction  of  the  2-axis  of  reference  towards 
the  reader  in  Fig.  104  we  have  from  equation  (1) : 

(a)  xY  is  an  x-vector  multiplied  by  a  ^/-vector,  and  it  is  a 
vector  in  the  positive  direction  of  the  2-axis. 

(6)  yX  is  a  ?/-vector  multiplied  by  an  a;-vector,  and  it  is  a  vector 
in  the  negative  direction  of  the  2-axis. 

Expressing  F  and  I  in  terms  of  their  components  as  shown  in 
Fig.  104,  we  have: 

F  =  X  +  y  (2) 

and 

l  =  x-\-y  (3) 

Multiplying  equations  (2)  and  (3)  member  by  member,  using  the 


VECTOR  ANALYSIS.  217 

ordinary  rules  of  algebra,  we  get : 

IF  =  xX -\- yY -{- xY -{-  yX  (4) 

But  it  is  shown  above  that  xY  and  yX  are  opposite  in  sign. 
Now  the  opposite  signs  oi  xY  and  yX  are  shown  in  equation  (4), 
according  to  the  above  discussion,  hy  the  fact  that  in  one  case  we 
have  an  x-vector  multiplied  by  a  y-vector  and  in  the  other  case  we  have 
a  y-vector  multiplied  hy  an  x-vector.  But  it  is  very  inconvenient 
to  have  algebraic  signs  so  indirectly  indicated;  it  is  better  to 
indicate  the  sign  explicitly  and  take  xY  —  yX  as  the  vector  part 
of  the  product  IF.  When  this  is  done  we  pay  no  attention  to  the 
order  of  the  factors,  because  the  inverse  order  in  the  two  terms  is 
taken  account  of  in  the  negative  sign  of  the  second  term.  With  this 
understanding,  therefore,  we  have  the  following  important  propo- 
sitions: 

(a)  The  scalar  part  of  the  product  IF  in  Fig.  104  is  equal  to 
xX  -h  yY.  This  is  evident  when  we  consider  that  xX  and  yY 
are  the  only  scalar  terms  in  equation  (4). 

(&)  The  vector  part  of  the  product  of  IF  in  Fig.  104  is  equal  to 
xY  —  yX,  and  it  is  in  the  direction  towards  the  reader  in  Fig.  104. 

116.  Scalar  fields. — It  is  sometimes  important  to  consider  the 
temperature  at  various  points  in  a  body,  the  pressure  at  various 
points  in  a  fluid,  the  density  at  various  points  of  a  substance  or 
the  electric  charge  per  unit  volume  at  various  points  in  a  region. 
Such  a  distribution  of  temperature,  hydrostatic  pressure,  or  density 
is  called  a  scalar  field  because  the  quantity  under  consideration  is 
a  scalar  and  it  has  a  definite  value  at  each  point  in  a  region  or 
field  of  space.  The  distribution  is  said  to  be  homogeneous  or 
uniform  when  the  scalar  quantity  has  the  same  value  at  every 
point  in  the  field;  otherwise  the  distribution  is  said  to  be  non- 
homogeneous.  An  example  of  a  non-homogeneous  scalar  field  is  the 
temperature  of  an  iron  rod  one  end  of  which  is  red  hot  and  the 
other  end  of  which  is  cold.  Atmospheric  pressure  is  also  non- 
homogeneously  distributed  because  it  is  different  at  different 
places  and  different  at  different  altitudes. 


218  CALCULUS. 

A  scalar  field  is  sometimes  called  a  distrihvied  scalar.  Thus 
when  the  scalar  nature  of  temperature  is  to  be  emphasized  it  is 
helpful  to  call  temperature  (throughout  a  hot  iron  rod,  for  example) 
a  distributed  scalar. 

A  scalar  field  is  sometimes  called  a  scalar  function,  A  distri- 
buted scalar  has  a  definite  value  at  each  point  of  space;  that  is, 
the  value  of  a  distributed  scalar  at  a  point  is  a  function  of  the 
coordinates  of  the  point.  When  this  fact  is  to  be  emphasized  it 
is  helpful  to  call  temperature  (or  any  distributed  scalar)  a  scalar 
function. 

117.  Vector  fields. — It  is  sometimes  important  to  consider  the 
velocity  at  different  points  of  a  fluid,  the  direction  and  intensity 
of  heat  flow  at  different  points  of  a  substance,  or  the  direction  and 
intensity  of  an  electric  or  magnetic  field  at  different  points  in 
space.  Such  a  distribution  of  fluid  velocity,  or  other  vector,  is 
called  a  vector  field  because  the  quantity  imder  consideration  is  a 
vector,  and  it  has  a  definite  value  and  direction  at  each  point  in 
a  region  or  field  of  space.  The  distribution  is  said  to  be  homo- 
geneoiLs  or  uniform  when  the  vector  has  the  same  value  and  is  in 
the  same  direction  at  every  point  in  the  field;  otherwise  the  distri- 
bution is  said  to  be  non-homogeneous.  The  water  in  a  rotating 
bowl  is  an  example  of  a  non-homogeneous  vector  field  because  the 
water  moves  in  different  directions  and  at  different  velocities  at 
different  points  in  the  bowl.  The  magnetic  field  around  an 
electric  wire  is  a  non-homogeneous  vector  field  because  the  mag- 
netic field  does  not  have  the  same  intensity  and  the  same  direction 
everywhere. 

A  vector  field  is  sometimes  called  a  distributed  vector.  Thus 
when  the  vector  nature  of  fluid  velocity  is  to  be  emphasized  it  is 
helpful  to  call  fluid  velocity  a  distributed  vector. 

A  vector  field  is  sometimes  called  a  vector  function.  Each 
component  (the  a;-component,  the  ^/-component  and  the  2-com- 
ponent)  of  the  velocity  of  a  fluid  has  a  definite  value  at  each  point 
of  space;  that  is,  the  values  of  the  components  at  a  point   p   are 


VECTOR  ANALYSIS.  219 

functions  of  the  coordinates  of  p.  When  this  fact  is  to  be  empha- 
sized it  is  helpful  to  call  fluid  velocity  (or  any  distributed  vector) 
a  vector  function. 

118.  Volume  integral  of  a  distributed  scalar. — For  the  sake  of 
simplicity  let  us  consider  a  special  case,  namely,  the  density  of  a 
substance,  and  let  us  assume  that  the  density  varies  from  point  to 
point.  Let  ^  be  the  density  at  a  given  point,  then  \l/  '  dr  is 
the  mass  of  material  in  the  volume  element  dr  at  the  point,  and 
the  total  mass  M  of  the  body  is: 

M  =  frP  '  dr  (1) 

This  integral  is  called  the  volume  integral  of  the  distributed  scalar  ^. 
The  physical  significance  of  volume  integral  is  not  in  every  case 
so  simple  as  in  the  case  of  density.  If  \l/  is  the  volume  density  of 
electric  ch?irge,  then  equation  (1)  gives  the  total  electric  charge 
in  the  region  throughout  which  the  integration  is  extended. 
If  ^  is  the  energy  density  in  an  electric  or  magnetic  field,  or  in  a 
strained  solid,  or  in  a  moving  fluid,  then  equation  (1)  gives  the 
total  energy  in  the  region  throughout  which  the  integration  is 
extended. 

119.  Gradient  of  a  distributed  scalar. — Any  distributed  scalar 
like  temperature,  or  density,  or  hydrostatic  pressure  has  a  definite 
value  at  each  point  of  a  field  and  therefore  the  value  of  any  distri- 
buted scalar  at  a  point  may  be  thought  of  as  a  function  of  the 
coordinates  x,  y  and  z  of  the  point.  Indeed,  this  matter  has 
already  been  discussed  in  Arts.  62  to  66,  where  it  is  explained 
that  a  distributed  scalar  has  a  definite  gradient  at  each  point. 
Thus  if  ^  is  the  temperature  at  a  point  in  a  body,  then 

(1) 

(2) 
(3) 


X  ■ 

dx 

Y 

dyp 

dy 

Z 

drP 

dz 

220  CALCULUS. 

where  X,  Y  and  Z  are  the  component  gradients  of  \f/.  The 
gradient  of  a  distributed  scalar  has  a  definite  value  and  a  definite 
direction  at  each  point  and  it  is,  therefore  a  distributed  vector. 

PROBLEMS. 

1.  A  vessel  one  meter  wide,  one  meter  long,  and  one  meter 
deep,  contains  a  fluid  of  which  the  density  is  one  gram  per  cubic 
centimeter  at  the  top,  increasing  uniformly  to  two  grams  per 
cubic  centimeter  at  the  bottom.  Find  the  volume  integral  of 
the  density  of  the  fluid.    Ans.  1,500,000  grams. 

2 .  The  gradient  of  the  density  in  problem  1  is  uniform  through- 
out the  vessel  and  equal  to  one  gram  per  cubic  centimeter  per 
meter,  and  it  is  directed  vertically  downwards.  Suppose  the 
downward  gradient  of  the  density  to  be  a  linear  function  of  the 
distance  from  the  top  of  the  vessel,  changing  from  zero  at  the  top 
to  two  grams  per  cubic  centimeter  per  meter  at  the  bottom. 
Find  the  volume  integral  of  the  density  of  the  fluid,  the  density 
being  one  gram  per  cubic  centimeter  at  the  top.  Ans.  1,333,333 
grams. 

3.  The  value  of  a  distributed  scalar  at  any  point   p   is   \p  =  — 

where  Q  is  a  constant  and  r  the  distance  of  p  from  the  origin 

of  coordinates.     Find  the  a:,  y  and  z  components  of  the  gradient 

Qx 
of  \l/.    Ans.  The  x-component  is    —  -r-r-; — t"; — 5Ti' 

{x^  +  2/2  H-  z^)* 

4.  Find  the  gradient  of  ^  (  =  ~ )    in  the  direction  of  r.    Ans. 

5.  The  a;-component  of  the  gradient  of  any  distributed  scalar 
may  be  thought  of  as  a  distributed  scalar.    Find  the  ^/-component 

of  the  gradient  of  the  a;-component  of  the  gradient  of  -•    Ans. 

SQxy 


(3^-\-y^  +  2^)1 


VECTOR  ANALYSIS.  221 

120.  Permanent  and  varying  states  of  scalar  distribution. — 

When  the  temperature  at  each  point  of  a  body  remains  unchanged, 
the  distribution  of  temperature  throughout  the  body  is  said  to  be 
permanent,  although,  of  course,  the  temperature  of  the  body 
may  not  be  everywhere  the  same.  When  the  temperature  at  each 
point  of  a  body  is  changing  we  have  what  is  called  a  varying  state 
of  distribution.  Thus  the  density  in  a  gas  which  is  being  com- 
pressed, and  the  temperature  throughout  a  body  which  is  being 
heated  or  cooled  are  examples  of  varying  scalar  distributions. 

121.  Stream  lines  of  a  distributed  vector. — A  line  drawn  through 
a  moving  fluid  so  as  to  be  at  each  point  in  the  direction  in  which 
the  fluid  at  the  point  is  moving  is  called  a  stream  line.  The 
geometrical  idea  of  a  stream  line  applies  to  any  vector  field  what- 
ever, and  the  manner  of  distribution  of  a  vector  is  clearly  repre- 
sented by  the  use  in  imagination  of  such  lines.  In  electric  and 
magnetic  fields  these  lines  are  called  lines  of  force,  but  the  term 
stream  line  will  be  used  in  general  statements. 

122.  Permanent  and  varying  states  of  vector  distribution. — 

When  the  velocity  of  a  moving  fluid  remains  unchanged  in  magni- 
tude and  direction  at  every  point,  we  have  what  is  called  a  perma- 
nent state  of  fluid  motion.  When  the  velocity  of  a  fluid  is  changing 
at  each  point,  we  have  what  is  called  a  varying  state  of  fluid  motion. 
Thus,  when  an  orifice  in  a  large  tank  of  water  is  suddenly  opened, 
a  perceptible  time  elapses  before  the  jet  of  water  becomes  estab- 
lished. During  this  time  the  velocity  of  the  water  is  changing 
rapidly  at  each  point  in  the  jet.  After  the  jet  becomes  steady, 
however,  the  velocity  of  the  water  at  each  point  remains  constant 
in  magnitude  and  in  direction.  The  magnetic  field  in  the  neigh- 
borhood of  a  moving  magnet  or  in  the  neighborhood  of  a  moving 
or  changing  electric  current  is  an  example  of  a  varying  vector 
distribution. 

Rate  of  change  of  a  distributed  vector  at  a  point. — Let  the  line  a, 
Fig.  105,  represent  the  value  at  a  given  instant,  of  the  velocity  of 
a  fluid  at  the  point    p,    and  let  the  line    a  +  Aa    represent  the 


222  CALCULUS. 

velocity  of  the  fluid  at  the  same  point  after  a  time-interval  A^  has 

elapsed.    The  Hmiting  value  oi  -r-   as  At  approaches  zero  is  the 

rate  of  change  of  a  at  the  point  p.*  This  rate  of  change  of  a 
has  a  definite  value  and  is  in  a  definite  direction  at  each  point  of 
space  and  it  is  therefore  a  distributed  vector. 

123.  Line  integral  of  a  distributed  vector. — Consider  a  line  or 
path  pp'  in  a  vector  field  as  shown  in  Fig.  106.  Let  As  be  an 
element  of  this  line  or  path,  let  R  be  the  value  of  the  distributed 


^direction  o/j^atp  N\ 

""\^.  ^^^ >P' 

Fig.  105.  Fig.  106. 

vector  at  the  element  As,  and  let  e  be  the  angle  between  U  and 
As.  Then  R  cos  €  is  the  resolved  part  of  R  parallel  to  As, 
and  R  cos  €  •  As  is  the  scalar  part  of  the  product  of  R  and  As; 
and: 

E  =  f  Rcose-  ds  (1) 

is  called  the  line  integral  of  the  distributed  vector  R  along  the  line 
or  path  over  which  this  summation  is  extended.  The  angle  e  is 
reckoned  between  R  and  the  positive  direction  of  As,  the  positive 

*  In  this  illustration  the  velocity  under  consideration  is  the  changing 
velocity  of  the  successive  particles  of  the  fluid  as  they  pass  the  point  p,  not 
the  changing  velocity  of  a  given  particle  while  it  is  traveling  along  near  p. 
The  latter  velocity  may  be  changing  from  instant  to  instant  even  though  the 
former  is  invariable. 


VECTOR  ANALYSIS.  223 

direction  of  As  being  the  direction  in  which  As  would  be  passed 
over  in  traveling  along  the  line  pp'  in  a  chosen  direction.  If  the 
chosen  direction  be  changed,  cos  e  will  change  sign  at  each  ele- 
ment. Therefore  the  line  integral  from  p  to  p'  is  equal  to  the 
line  integral  from  p'   to  p  in  Fig.  106,  but  opposite  in  sign. 

Examples. — The  line  integral  of  electric  field  along  a  path  is 
called  the  electromotive  force  along  the  path.  The  Une  integral  of 
a  magnetic  field  along  a  path  is  called  the  magnetomotive  force 
along  the  path.  The  line  integral  of  fluid  velocity  along  a  path 
is  called  the  circulation  of  the  fluid  along  the  path. 

Cartesian  expression  for  line  integral. — Let  X,  Y  and   Z  be 

the  components  of  the  vector  R,  and  let  dx,  dy  and  dz  be  the 
components  of  the  line  element   ds.     Then  we  have: 

R  =  X  +Y  +  Z     (a  vector  equation)  (2) 

and 

ds  =  dx  +  dy  +  dz     (a  vector  equation)  (3) 

The  product  of  the  two  vectors  R  and  ds  is  part  scalar  and 
part  vector  as  explained  in  Arts.  114  and  115,  and  the  scalar  part 
of  the  product  is  R  cos  e  -  ds  or  X  -  dx  +  Y  -  dy  +  Z  •  dz. 
Therefore  equation  (1)  may  be  written: 

E  =  J{X  'dx+Y  'dyi-Z  -dz)  (4) 

Line  integral  of  the  gradient  of  a  distributed  scalar.— Consider 
a  distributed  scalar  rf/.  Let  us  call  it  temperature  for  the  sake 
of  inteUigibility.  Let  R  in  Fig.  106  be  the  gradient  of  \p,  that 
is  R  is  the  temperature  gradient.  Then  the  line  integral  of  R 
along  the  path  pp'  is  equal  to  \J/'  —  ^,  where  ^'  is  the  tempera- 
ture at  p'  and  \f/  is  the  temperature  at  p.  This  is  evident 
from  the  following  considerations.  The  product  R  cos  €  in  Fig. 
106  is  the  resolved  part  of  the  temperature  gradient  in  the  direction 
of  the  line  element  As,  and  i^  cos  €  •  As  is  the  change  of  tempera- 
ture along  As.  Therefore  2/2  cos  e  •  As  is  the  sum  of  the  changes 
of  temperature  along  all  parts  of  the  path  pp'  or  the  total  change 
of  temperature  from  p  to  p'. 


224 


CALCULUS. 


Line  integral  of  a  gradient  along  co-terminous  paths. — The 
line  integral  of  R  along  any  path  pp'  is  the  temperature  difference 
between  p  and  p'.  Therefore  the  line  integral  of  R  is  the  same 
for  all  paths  from  p  to  p\  There  is  an  important  exception  to 
this  proposition  as  follows: 

Imagine  an  ordinary  auger  to  be  placed  with  its  axis  at  right 
angles  to  the  plane  of  the  paper  in  Fig.  107,  the  plane  of  the 


Fig.  107. 
Slope-lines  on  an  auger-hill. 


Fig.  108. 

Magnetic  lines  of  force  around  an 
electric  wire. 


paper  being  the  base  plane  and  the  winding-stair-like  surface  of 
the  auger  being  looked  upon  as  the  surface  of  a  hill  raised  above  the 
base  plane.  This  hill  we  will  call  an  auger-hill,  and  its  slope 
lines  are  shown  by  the  fine-line  circles  in  Fig.  107.  The  height 
of  this  auger-hill  above  any  point  p  has  many  values  differing 
from  each  other  by,  say,  one  inch,  where  one  inch  is  the  ** pitch" 
of  the  auger. 

Now  the  line  integral  of  the  slope  of  a  hill  along  any  path  is  the 
difference  of  level  of  the  ends  of  the  path.  A  path  which  returns 
to  its  starting  point  (a  closed  path)  comes  back  to  its  initial  level 
on  an  ordinary  hill,  but  a  path  from  p  back  to  p  in  Fig.  107  does 
not  come  back  to  its  initial  level  if  the  path  circles  round  the  axis 
of  the  auger.  Therefore  the  line  integral  of  slope  along  the 
path  1  from  p  to  p'  in  Fig.  107  is  not  the  same  as  the  line  integral 
of  slope  along  path    2    from  p  to  p'. 


VECTOR  ANALYSIS.  225 

The  fine-line  circles  in  Fig.  108  are  the  magnetic  lines  of  force 
(slope  lines  of  the  magnetic-potential  hill)  in  the  neighborhood 
of  an  electric  wire  perpendicular  to  the  plane  of  the  paper.  The 
line  integral  of  the  magnetic-potential  slope  along  path  1  from 
p  to  p'  in  Fig.  108  is  not  the  same  as  the  line  integral  of  the 
magnetic-potential  slope  along  path    2    from  p  to  p'. 

124.  Potential  of  a  vector  field. — Consider  any  distributed  vec- 
tor R,  Then  the  '^height"  at  each  point  of  space  of  an  imagined 
"hill'' whose  slope  or  gradient  is  everywhere  equal  to  R  is  called 
the  potential  of  R.  Thus  the  temperature  at  each  point  of  a 
body  is  the  "height"  at  that  point  of  a  hill  whose  slope  is  every- 
where equal  to  the  temperature  gradient,  therefore  temperature 
is  the  potential  of  temperature  gradient.  The  "height"  in  volts 
at  each  point  of  space  of  an  imagined  "hill"  whose  slope  is  every- 
where equal  to  the  electric  field  intensity  (in  volts  per  centimeter) 
is  called  electric  potential.  The  "height"  at  each  point  of  space 
of  an  imagined  "hill"  whose  slope  is  everywhere  equal  to  the 
velocity  of  a  moving  fluid  is  called  the  velocity  potential  of  the  fluid. 

The  temperature  at  a  point  of  a  body  can  of  course  be  deter- 
mined by  placing  a  thermometer  at  that  point.  That  is,  tempera- 
ture is  an  actual  physical  condition.  The  electric  potential  at 
a  point,  however,  is  not  an  actual  measurable  physical  condition 
at  that  point;  indeed  electric  potential  exists  only  in  the  imagi- 
nation as  a  sort  of  mathematical  fiction  whose  usefulness  grows 
out  of  the  fact  that  it  is  often  very  helpful  to  think  of  a  given 
vector  field  as  a  gradient.  An  example  showing  the  usefulness 
of  the  idea  of  potential  is  given  in  Art.  133. 

Theorem  as  to  the  existence  of  potential. — Let  ^  be  a  distri- 
buted scalar,  like  temperature.  Then  the  component  gradients 
of  yp  are  given  by  equation  (1),  (2)  and  (3)  of  Art.  119.  Differ- 
entiating the  first  of  these  equations  with  respect  to  y  and  the 
second  with  respect  to  x  we  have: 

d^yp         dX 


dy  '  dx       dy 


16 


226  CALCULUS, 

and 


^         dY  (2^ 


dx  '  dy      dx 

But   J       ,     and    ,       ,     are   always  identical   as  stated  in 
dy  '  dx  dx  '  dy 

Art.  59  and  as  demonstrated  in  Art.  134.    Therefore  from  equa- 
tions (1)  and  (2)  we  have: 

^  ^dY 

dy       dx 

Similarly  from  equations  (2)  and  (3)  of  Art.  119,  we  obtain: 

f-^  =  0  (4) 

dz       dy 

and  frona  equations  (1)  and  (3),  we  obtain: 

dZ      dX  ,  V 

In  these  equations  (3),  (4)  and  (5)  X,  Y  and  Z  are  the  com- 
ponent gradients  of  any  scalar  function  \J/,  and  any  distributed 
vector  whose  components  satisfy  equations  (3),  (4)  and  (5)  can  have 
a  potential,  or,  in  other  words,  any  distributed  vector  whose  com- 
ponents satisfy  (3),  (4)  and  (5)  can  be  looked  upon  as  the  gradient 
of  an  imagined  "hilV 

Multivalued  potential. — The  auger-hill  which  is  represented  in 
Fig.  107  has  a  multiplicity  of  heights  above  any  point  p.  Similarly 
the  magnetic-potential  in  Fig.  108  has  a  multiplicity  of  values  at 
any  point  p.  Of  course  the  "height"  of  the  potential  "hill" 
in  Fig.  108  may  be  thought  of  as  an  actual  height  measured 
upward  from  the  plane  of  the  paper;  but  one  must  not  forget  that 
the  magnetic  field  under  consideration  fills  all  space,  and  that 
there  is  a  multiplicity  of  values  of  magnetic  potential  at  each 
point  in  space.    See  Art.  66. 


VECTOR  ANALYSIS. 


227 


125.  Surface  integral  of  a  distributed  vector.  Flux. — 
Let  AOOB,  Fig.  109,  be  a  diaphragm  stretching  across  a  closed 
loop  of  wire,  the  dots  A  and  B  being  where  the  wire  passes 
through  the  plane  of  the  paper.    Let   AS   be  the  area  of  an  ele- 


diaphram 


Fig.  109. 


Fig.  110. 


ment  of  the  diaphragm,  let  R  be  the  value  at  AS  of  a  distributed 
vector,  and  let  e  be  the  angle  between  R  and  the  normal  to  AS 
this  normal  being  always  drawn  outward  from  the  same  side  00 
of  the  diaphragm.  Then  R  cos  e  is  the  resolved  part  of  R  normal 
to  A>S,  and  R  cos  e  '  AS  is  the  scalar  part  of  R  -  AS;  and: 


^  =  f  Rcose  '  dS 


(1) 


is  called  the  surface  integral  of  R  over  the  portion  of  the  diaphragm 
over  which  the  integration  is  extended. 

^  If  the  normal  to  AS  in  Fig.  109  is  reversed,  as  shown  in  Fig 
110,  then  cos  e  will  be  reversed  in  sign,  and  the  surface  integral, 
retaining  its  numerical  value,  will  be  reversed  in  sign. 

In  the  integration  over  a  closed  surface  like  a  box  or  sphere, 
the  normal  is  understood,  throughout  the  following  discussion, 
to  be  drawn  outwards. 

Examples. — If  R  in  Fig.  109  is  the  velocity  of  a  fluid  at  AS, 
then   R  cos  e   is  the  resolved  part  of  the  velocity  normal  to   AS, 


228  CALCULUS. 

R  cos  c  •  A*S  is  the  volume  of  fluid  per  second  flowing  across 
AS,  and  JR  cos  e  •  dS  is  the  total  volume  of  fluid  per  second 
flowing  through  the  loop  of  wire  AB.  The  volume  of  fluid  per 
second  passing  through  a  loop  is  called  the  fliix  of  the  fluid  through 
the  loop,  and  the  surface  integral  of  any  distributed  vector  over  a 
surface  is  called  the  fiux  of  the  vector  across  the  surface.  Thus 
magnetic  flux  is  the  surface  integral  of  magnetic  field  and  electric 
flux  is  the  surface  integral  of  electric  fleld. 

The  surface  integral  of  a  distributed  vector  over  a  closed  surface 
like  a  box  or  sphere  is  called  the  flux  into  or  oiU  of  the  region 
enclosed  by  the  box  or  sphere. 

Cartesian  expression  for  surface  integral. — Let   X,  Y   and   Z 

be  the  componenfs  of  the  vector  R  in  Fig.  109,  and  let  da,  db 
and  dc  be  the  areas  of  the  projections  of  dS  on  the  yz,  on  the 
xz,   and  on  the  xy  planes,  respectively.     Then: 

R  =  X  +  Y  +  Z    (a  vector  equation)  (2) 

and 

dS  =  da  -{-  dh  +  dc     (a  vector  equation)  (3) 

The  scalar  part  of  the  product  R  -  dS  is 

X  '  da  +  Y  '  dh  -\-  Z  '  dc  i=  R  cos  €  '  dS), 

But  the  shapes  of  the  surface  elements  da,  db  and  dc  may  be 
anything  whatever.  Therefore  we  may  write  dy  •  dz  for  da, 
dx  •  dz  for  dh,  and  dx  •  dy  for  dc.    Hence  equation  (1)  becomes: 

^  =  ff  (X  '  dy  '  dz  -{-  Y  '  dx  '  dz  +  Z  '  dx  '  dy)        (4) 

126.  Divergence  of  a  distributed  vector. — In  some  cases  a  dis- 
tributed vector  "flows"  outwards  or  emanates  from  a  region  of 
space.  Thus  the  liquid  in  a  tank  flows  outwards  from  the  end 
of  a  supply  pipe.  When  a  gas  is  expanding  each  portion  of  the 
gas  is  growing  less  dense  and  there  is  an  outward  flow  from  every 
small  part  of  the  region  occupied  by  the  expanding  gas.  What  is 
called  electric  field  emanates  from  electrically  charged  bodies, 


VECTOR  ANALYSIS.  229 

and  when  electric  charge  is  spread  throughout  a  region,  electric 

field  emanates  from  every  small  part  of  the  region  occupied  by 

the  charge. 

Consider  a  small  volume  element   Ar  in  the  neighborhood  of  a 

point  p  in  a  vector  field.     Let  A$  be  the  flux  of  the  distributed 

vector    R    out  of  the  volume  element    Ar.     It  can  be  shown,* 

when  i2  is  a  continuous  function  of  the  coordinates  x,  y  and  2;, 

A$ 
that  the  ratio    7-     approaches  a  definite  limiting  value  as  the 

At 

volume  element    Ar    grows  smaller  and  smaller.    This  limiting 

value,   -J-    is  called  the  divergence  of  the  vector  field  at  the  point  p. 

Therefore,  representing  the  divergence  of  a  vector  field  at  a  point 
by  p,  we  have: 

d^  =  P'  dr  (1) 

in  which  d^  is  the  flux  of  R  coming  out  of  the  volume  element  dr. 

When  a  vector  field  flows  into  each  small  part  of  a  region,  the 
divergence  is  negative.  Negative  divergence  is  sometimes  called 
convergence. 

The  flux  $  of  a  vector  field  across  a  surface  is  a  scalar  quantity, 
as  is  evident  from  the  discussion  of  equation  (1)  of  Art.  125. 

(A<l>\ 
—  1 

is  a  scalar  quantity;  and  since  a  vector  field  has  a  definite  diver- 
gence at  each  point  (of  course  the  divergence  may  be  zero)  of 
space,  it  is  evident  that  the  divergence  of  a  vector  function  (a 
distributed  vector)  is  a  scalar  function  (a  distributed  scalar). 

Cartesian  expression  for  divergence. — Consider  a  small  cube  of 
which  the  edges  are  dx,  dy  and  dz  as  shown  in  Fig.  111.  Let 
Xj  Y  and  Z  be  the  components  of  the  given  distributed  vector  R 

A€» 
*  The  actual  proof  that  —  has  a  definite  limiting  value  may  be  estab- 
lished without  great  difficulty  by  considering  the  portion    v"    of  the  linear 
vector  field  in  Art.  133. 


230 


CALCULUS. 


at  the  point  p.  Then  the  flux  of  R  into  the  cube  across  the 
face  a  is  X  *  dy  '  dzj  because  X  is  the  component  of  R  normal 
to  the  face   a   and   dy  •  dz   is  the  area  of  face   a.    If   X   is  the 

ic-component  of  R  at  any  point  of  the  face  a,  then  f  X  +  -7-  *  dx\ 


z-axis 

^y^r- 

p^ 

JL. 

^^#^ 

1 

1 
\' 

1 

X-axis 

yy-axi 

B 

X 

1 

/ 

dx 


Fig.  111. 

is  the  a;-component  of  R  at  the  corresponding  point  of  the  face  6, 
and  if   X  -  dy  -  dz   is  the  flux  into  the  cube  across  face   a   then 

X  +  ^  '  dx)  '  dy  '  dz  is  the  flux  out  of  the  cube  across  face  &.* 

Therefore  the  flux  of  R  out  of  the  cube  across  face  h  exceeds  the 

jy 

flux  of  R  m^o  the  cube  across  face  a  by  the  amount  -7-  -dx'dy'dz. 

In  the  same  manner  it  may  be  shown  that  the  net  flux  of  R  out  of 
the  cube  across  the  two  faces  which  are  perpendicular  to  the  y-axis 

is  -T~  '  dx  •  dy  '  dz;   and  that  the  net  flux  oui  of  the  cube  across 

the  two  faces  which  are  perpendicular  to  the  2-axis  is  -r  'dx-dy-  dz. 
Therefore  the  total  flux  out  of  the  cube  is: 


,^       /dX,dY^dZ\      ,       ,       , 
^^=Kdx-^dy^-dz)'  <^^'dy'dz 


(2) 


*The  argument  here  given  is  not  entirely  rigorous,  and  the  peculiar 
wording  of  this  sentence  is  intended  to  suggest  a  form  of  argument  which  is 
rigorous. 


VECTOR  ANALYSIS.  231 

But  dx  •  dy  '  dz  is  the  volume  dr  oi  the  cube.    Therefore,  using 
equation  (1)  we  have: 

.  dX.dY.dZ  .^. 

PROBLEMS. 

1.  All  space  is  filled  with  a  fluid  moving  parallel  to  the  a;-axis 
of  reference  at  a  uniform  velocity  of  10  centimeters  per  second. 
Find  an  expression  for  the  velocity  potential  of  the  fluid.  Ans. 
xp  =  10a;  +  any  constant. 

2.  The  velocity  components  of  a  moving  fluid  parallel  to  the 
axes  of  reference  are  everywhere  equal  to  a,  h  and  c  respectively. 
Find  the  velocity  potential.  Ans.  rj/  =  ax  +  hy  -\-  cz  +  any  con- 
stant. 

3.  The  velocity  components  of  a  fluid  parallel  to  the  axes  of 
reference  are  ax,  hy  and  cz,  respectively.  Find  the  velocity 
potential.     Ans.  \J/  =  \ax^  +  \hy'^  +  \cz^  +  any  constant. 

4.  A  viscous  fluid  flowing  over  a  plane  has  a  velocity  which  is 
everywhere  given  by  the  equation  X  =  ay  as  shown  in  Fig.  p4. 
Show  that  no  velocity  potential  exists. 


y-axi8 


*■  i  ,>  X  a*  gj/ 


Jub:. 


x-axi8 


Fig.  p4. 

5.  A  vessel  of  water  rotates  uniformly  at  a  speed  of  two  revo- 
lutions per  second  about  a  vertical  axis  (the  2-axis).  Derive 
expressions  for  the  velocity  components  of  the  moving  water  and 
show  that  the  velocity  has  no  potential. 

6.  Liquid  flows  over  an  infinite  plane  towards  a  circular  spot 


232  CALCULUS. 

20  centimeters  in  radius  where  the  liquid  leaks  through  the  plane 
at  the  rate  of  two  cubic  centimeters  per  second  for  each  square 
centimeter  of  the  leaky  portion  of  the  plane.  The  depth  of  the 
liquid  is  everywhere  10  centimeters.  Choose  the  x  and  2/-axes 
of  reference  in  the  plane  with  the  origin  at  the  center  of  the  leaky 
portion,  and  derive  expressions  for  the  x  and  2/-components  of 
the  fluid  velocity  in  the  region  over  the  leaky  portion.    Ans.  x- 

component  =  r^:. 

Note. — In  this  and  the  following  problems  consider  only  the  horizontal 
part  of  the  motion  and  assume  the  velocity  to  be  the  same  from  top  to  bottom 
of  the  liquid. 

7.  Find  an  expression  for  the  velocity  potential  in  the  region 

x^  -I—  ifi 
over  the  leaky  portion.    Ans.  — ^ — h  any  constant. 

8.  Derive  expressions  for  the    x    and    y    components  of  the 

fluid  velocity  in  the  region  outside  of  the  leaky  portion.    Ans.  x- 

40x 
component  =  -^x~2- 

9.  Find  an  expression  for  the  velocity  potential  in  the  region 
outside  of  the  leaky  portion.    Ans.  20  logc  (x^  -f  y"^)  +  any  constant. 

10.  Determine  the  constants  of  integration  involved  in  the 
answers  to  problems  7  and  9  on  the  assumption  that  the  velocity 
potential  is  zero  at  the  edge  of  the  leaky  spot.  Ans.  (a)  —  20, 
(6)  -  20  log.  400. 

11.  A  straight  "fence"  20  centimeters  long  and  10  centimeters 
high  is  placed  in  the  layer  of  moving  liquid  on  the  plane  in 
problem  6.  Find  the  surface  integral  of  the  fluid  velocity  over 
the  fence  (a)  when  the  ends  of  the  fence  are  at  a  distance  of  Vl25 
centimeters  from  the  center  of  the  leaky  spot,  and  (6)  when  the 
ends  of  the  fence  are  V  1,000  centimeters  from  the  center  of  the 
leaky  spot.  Ans.  (a)  100  cubic  centimeters  per  second,  (6)  800 
tan~i  0.333  cubic  centimeters  per  second. 

12.  Find  the  divergence  of  the  fluid  velocity  in  problem  6  (a) 
over  the  leaky  spot  and  (6)  in  the  region  outside  of  the  leaky  spot. 


VECTOR  ANALYSIS. 


233 


Ans.  (a)  0.2  cubic  centimeter  per  second  per  cubic  centimeter,  (h) 
zero. 

13.  Find  the  divergence  of  the  fluid  velocity  specified  in  prob- 
lem 3.    Ans.  a  -f-  6  +  c; 

14.  One  end  of  a  rubber  band  is  fixed,  and  the  band  is  stretched 
by  moving  the  other  end  of  the  band  at  a  velocity  of  2  centimeters 
per  second.  The  band  is  24  centimeters  long  at  a  given  instant. 
Find  the  divergence  of  v  where  v  is  the  distributed  velocity  of 
the  band.  Assume  that  the  band  does  not  contract  laterally.  Ans. 
0.0833  cubic  centimeters  per  second  per  cubic  centimeter. 

15.  Find  the  distribution  of  pressure  in  a  tank  of  water  rotating 
at  a  speed  of  one  revolution  per  second  (w  =  27r  radians  per 
second),  the  density  of  water  being  62.5  pounds  per  cubic  foot. 
Ans.  p  =  123.3  r^  where  p  is  the  pressure  in  poundals  per  square 
foot  at  a  point  r  feet  from  the  axis  of  rotation. 

Note. — Consider  an  element  of  the  rotating  liquid  as  indicated  by  the 
shaded  area  in  Fig.  pl5,  the  dimen- 
sion perpendicular  to  the  plane  of  the 
paper  being  I.  The  area  of  side  a  is 
Ir  .  do  and  the  outward  push  on  a  is 
plr  .  do.  The  area  of  the  side  h  is 
Z(r  4-  dr)  .  do,  and  the  pressure  at  this 
face  is  (p  +  dp) ;  therefore,  dropping 
infinitesimals  of  the  third  order,  the 
force  pushing  inwards  on  face  h  is  plr 
.  do  -{-  pl-dr  .do  +lr  .pd.dd.  The 
area  of  faces  c  and  e  is  I  .  dr  and 
the  pressure  over  these  faces  may  be 
considered  as  equal  to  p;  therefore  the 
normal  forces  on  faces  c  and  e  are 
pl.dr  as  shown.  The  outward  compo- 
nent of  these  two  forces,  according  to 


Fig.  pl5. 


Art.  51,  is 


pi  .  dr 


Xr 


(omitting   infinitesimals   of   the   third   order). 


Therefore  the  net  inward  force  due  to  pressure  acting  on  the  element  of  liquid 
is  Ir  .  dp  .  de,  and  this  must  be  equal  to  the  product  of  the  mass  Dlr  .  dd  .  dr 
of  the  element  of  liquid  and  its  radial  acceleration  coV. 

16.  Find  the  expressions  for  the  pressure  gradients  in  the 


234  CALCULUS. 

rotating  liquid  of  problem  15  in  the  directions  of  x  and  y  axes 
lying  in  a  plane  at  right  angles  to  the  axis  of  rotation.  Ans. 
X  =  246.6X,  Y  =  246.6y. 

17.  Find  the  divergence  of  the  pressure  gradient  in  the  rotating 
fluid  of  problem  15.    Ans.  493.2. 

127.  Reduction  of  a  surface  integral  over  a  closed  surface  to  a 
volume  integral  extended  throughout  the  enclosed  region. — 
A  very  important  transformation  in  the  theory  of  electricity  and 
magnetism  is  the  reduction  of  a  surface  integral  to  a  volume 
integral  or  vice  versa.  Let  J?  be  a  distributed  vector,  and  let 
us  consider  its  surface  integral  over  a  closed  surface  (normal 
directed  outwards).  Imagine  the  entire  enclosed  region  to  be 
broken  up  into  small  cells  Uke  the  individual  bubbles  in  a  mass  of 
foam.  Then  the  integral  of  R  over  the  hounding  surface  of  the  region 
is  equal  to  the  sum  of  the  integrals  of  R  over  the  hounding  surfaces  of 
the  individuxil  cells  (normal  directed  outwards  in  each  case).  This 
proposition  is  evident  when  we  consider  that  every  wall  which 
separates  two  cells  is  integrated  over  twice  with  directions  of 
normal  opposite  (see  Art.  125),  so  that  the  surface  integrals  over 
dividing  walls  are  thus  cancelled.  The  only  surface  integrals 
which  are  not  thus  cancelled  are  the  integrals  over  the  parts  of 
the  surface  which  bounds  the  region. 

Let  d$  be  the  surface  integral  over  one  of  the  cells  (the  flux 
of  R  out  of  the  cell),  and  let  Jr  cos  e  -  dS  be  the  surface 
mtegral  of  R  over  the  bounding  surface  of  the  enclosed  region. 
Then  from  the  above  proposition  we  have; 

fR  cos  €'  dS  =  fd^  (1) 

But  d^  =  p  '  dr  according  to  equation  (1)  of  Art.  126,  so  that 
equation  (1)  becomes: 

Jr  cos  €  •  dS  =  fp  '  dr  (2) 

This  equation  may  be  expressed  in  words  as  follows :  The  integral 
of  a  distributed  vector  R  over  a  closed  surface  is  equal  to  the  integral 


VECTOR  ANALYSIS.  235 

of  the  divergence  of  R  throughout  the  enclosed  region.  The  first 
member  of  (2)  is  of  course  a  surface  integral  and  the  second 
member  is  a  volume  integral. 

128.  Solenoidal  vector  fields. — A  vector  field  is  said  to  be 
solenoidal  when  its  divergence  is  everywhere  zero.  The  surface 
integral  of  such  a  vector  field  over  any  closed  surface  is  zero 
according  to  equation  (2)  of  Art.  127. 

Tube  of  flow. — Imagine  stream  lines  to  be  drawn  in  a  vector 
field  from  each  point  of  the  periphery  of  a  closed  curve  or  loop. 
These  stream  lines  form  a  tubular  surface  which  is  called  a  tube 
of  flow  of  the  given  distributed  vector.  Consider  a  number  of 
diaphragms  which  stretch  across  a  tube  of  flow.  The  flux  is  the 
same  across  each  diaphragm.  This  is  evident  when  we  consider 
that  any  two  diaphragms,  together  with  the  walls  of  the  tube, 
constitute  a  closed  surface  out  of  which  the  total  flux  must  be 
zero  because  the  distributed  vector  under  consideration  is  assumed 
to  be  solenoidal;  but  the  flux  across  the  walls  of  the  tube  is  zero 
because  the  vector  field  is  everywhere  parallel  to  the  walls.  There- 
fore the  flux  into  the  enclosed  space  across  one  diaphragm  must 
be  equal  to  the  flux  out  of  the  enclosed  space  across  the  other 
diaphragm. 

Unit  tube. — A  tube  of  flow  is  called  a  unit  tube  when  the  flux 
through  the  tube  is  unity.  Thus,  for  example,  a  unit  tube  has  a 
sectional  area  of  0.1  of  a  square  inch  at  a  point  in  a  moving  fluid 
where  the  velocity  of  the  fluid  is  10  inches  per  second.  A  unit 
tube  has  a  sectional  area  of  0.1  of  a  square  centimeter  at  a  point 
in  a  magnetic  fleld  where  the  intensity  of  the  field  is  10  gausses. 
Imagine  the  solenoidal  region  of  a  distributed  vector  to  be  divided 
up  into  unit  tubes.  Then  the  flux  across  any  surface  anywhere 
in  the  region  will  be  equal  to  the  number  of  these  unit  tubes  which 
pass  through  the  surface.  Each  unit  tube  may  be  conveniently 
represented  in  imagination  by  the  single  stream  line  along  the  axis 
of  the  tube.  When  stream  lines  are  drawn  in  this  way,  so  that 
each  stream  line  represents  a  unit  tube,  then  the  flux  across  any 


236  CALCULUS. 

surface  in  the  vector  field  is  equal  to  the  number  of  stream  Unes 
which  pass  through  the  surface.  The  lines  of  force  in  a  magnetic 
or  electric  field  are  always  thought  of  as  being  drawn  so  that  each 
line  represents  a  unit  tube,  and  the  quantity  of  magnetic  flux  or 
electric  flux  through  a  surface  is  expressed  by  the  number  of  lines 
of  force  which  pass  through  the  surface. 

129.  Curl  of  a  distributed  vector. — In  Art.  126  examples  were 
given  of  vector  fields  which  emanate  or  "flow  out"  from  a  region 
or  from  every  small  part  of  a  region.  There  are  important  cases 
in  which  a  vector  field  curls  round  a  region  or  round  every  small 
part  of  a  region.  Thus  the  stream  lines  in  a  rotating  bowl  of  water 
curl  round  the  central  portions  of  the  bowl.  The  magnetic  field 
due  to  an  electric  wire  curls  round  the  wire.  The  electric  field 
induced  by  an  iron  rod  while  the  rod  is  being  magnetized  curls 
round  the  rod. 

Consider  a  small  plane  area  AS  at  a  point  p  in  a.  vector  field. 
Let  AL  be  the  fine  integral  of  the  distributed  vector  R  round  the 
boundary  of  this  element  of  area.  It  can  be  shown,*  when  R  is 
a  continuous  function  of  the  coordinates    Xy  y    and    z,    that  the 

ratio  -r-s  approaches  a  definite  limiting  value  as  the  element 

of  area   AS  grows  smaller  and  smaller.    This  limiting  value,   -r^ 

is  the  component  of  a  new  vectorf  C  in  the  direction  of  the  normal 
to  dSf  and  this  new  vector  C  is  called  the  curl  of  the  given 
vector   R   at  the  point   p.    From  this  definition  we  have: 

dL  =  C  cose-  dS  (1) 

where    dL    is  the  line  integral  of    R    round  the  boundary  of  a 

*  The  actual  proof  that   — „  approaches  a  definite  limiting  value  may  be 

established  without  great  difficulty  by  considering  the  part   v'   of  the  linear 
vector  field  in  Art.  133;  see  Art.  135. 
t  See  Art.  135. 


VECTOR  ANALYSIS.  237 

small  plane  element  of  area  dS,  C  is  the  curl  of  R  at  the  element 
of  area,  and  e  is  the  angle  between  C  and  the  normal  to  dS. 

Example  of  curl. — Consider  a  uniformly  rotating  bowl  of  water, 
the  speed  of  rotation  being  oi  radians  per  second.  Consider  the 
character  of  the  fluid  motion  in  the  immediate  neighborhood  of 
the  point  p  distant  D  from  the  axis  of  rotation  of  the  bowl  as 
shown  in  Fig.  112.  The  motion  of  a  small  portion  of  the  water 
near  p  may  be  thought  of  as  a  combination  of  (1)  A  motion  of 
translation  at  velocity  Deo,  and  (h)  A  simple  motion  of  rotation 
at  a  speed  of  w  radians  per  second  about  an  axis  through  the 
point  p.*  Now  in  considering  the  line  integral  of  the  fluid 
velocity  around  the  circle  cc,  the  motion  of  translation  evidently 
need  not  be  considered.  Let  r  be  the  radius  of  the  circle  cc. 
The  velocity  of  the  fluid  at  every  point  of  the  circumference  of  cc 
(rotatory  motion  only  being  considered)  is  cor  and  it  is  everywhere 
in  the  direction  of  the  circumference.  There- 
fore the  angle  €  is  zero  and  the  expression 
for  the  line  integral  around  the  circle  reduces 
to  circumference  X  (>ir.  Therefore  AL  = 
27rr2aj,  where  AL  is  the  line  integral  of  the 
fluid  velocity  around  the  circle  cc.  But  the 
area  of  the  circle  is  dS  =  ttt^.  Therefore  from 
equation  (1)  we  get:  C  =  2aj.  That  is,  the 
curl  of  the  fluid  velocity  in  the  rotating  bowl  is  Yig.  112. 

equal  to  two  times  the  speed  of  the  bowl  in  radi- 
ans per  second;  and  since  the  expression  for  C  does  not  contain   D 
it  is  evident  that   C  has  the  same  value  everywhere  in  the  bowl. 

Now  the  angular  velocity  co  of  the  bowl  is  a  vector  and  its 
vector  direction  is  parallel  to  the  axis  of  rotation  and  towards  the 
reader  in  Fig.  112.  Therefore  C  is  a  vector,  and  it  is  towards  the 
reader  at  every  point  in  Fig.  112.  The  "stream  lines"  of  the 
vector  C  are  straight  lines  parallel  to  the  axis  of  rotation  of  the 
bowl. 

*  See  Franklin  and  MacNutt's  Mechanics  and  Heat,  pages  174-175,  The 
Macmillan  Co.,  New  York,  1910. 


238 


CALCULUS. 


Cartesian  expression  for  curl. — Consider  a  small  rectangular 
plane  area  of  which  the  edges  are  dy  and  dz,  as  shown  in  Figs. 
113  and  114.    Let  X,  Y  and  Z  be  the  components  at  p  of  the 


y-axia 


z-axU 


y-axi$ 


Q 


Z•^^'f 


z  ^ 

z-axia 


Fig.  113. 


Fig.  114. 


given  distributed  vector  R.  To  find  the  line  integral  of  R  around 
the  small  rectangle  in  the  direction  of  the  small  curved  arrow,  let 
us  consider  first  th^  two  edges  a  and  6  in  Fig.  113.  The  line 
integral  along  the  edge  a  is  —  Y  •  dy  because  the  component  Y 
is  counter  to  the  direction  of  the  small  curved  arrow.  If  Y  is 
the  i/-component  of  R  at  any  point  of  the  edge  a,  then  the 
2/-component  of    R    at  the  corresponding  point  of  the  edge    h 

dz;  and  if  the  line 


being   -3-  •  dz   greater,  is  equal  to    F  +    , 
az  dz 


integral  of  R  along  edge  a  is  —  Y  '  dy,  then  the  line  integral 
of  R  along  the  edge  h  is  +  (y  -{-  -^-dzj-dy*  Therefore  the 
net  value  of  the  line  integral  along  the  two  edges  a  and  b  is 
-f  '  dy  »  dz.  In  the  same  way  it  may  be  shown  from  Fig.  114 
that  the  net  line  integral  (in  the  direction  of  the  small  curved 
arrow)  along  the  edges   e   and  /is    —  -r-  '  dy  -  dz.    Therefore 


dZ     , 

-Ty'^y 


*  The  argument  here  given  is  not  entirely  rigorous,  and  the  peculiar  wording 
of  this  sentence  is  intended  to  suggest  a  slightly  modified  argument  which  is 
rigorous. 


VECTOR  ANALYSIS.  239 

the  total  line  integral  dL  of  R  around  the  small  rectangle  is: 

But  dy  '  dz  is  the  area  dS  of  the  rectangle.  Therefore  using 
equation  (1)  and  remembering  that  the  x-component  of  the  curl 
is  at  right  angles  to  the  rectangle  in  Figs.  113  and  114,  we  have: 

^^-  dz~  dy  ^^^ 

In  a  similar  manner  we  may  get: 

^'~  dx      dz  ^^^ 


and 


^-g-S  («) 


where  d,  Cy  and  C  z  are  respectively  the  x^  y  and  z  com- 
ponents of  the  curl  of  the  distributed  vector  R  whose  x,  y  and  z 
components  are  X,  Y  and   Z  respectively. 

The  curl  of  a  gradient  is  necessarily  equal  to  zero. —  This  is 
evident  when  we  consider  that  equations  (3),  (4)  and  (5)  of  Art. 
124  are  satisfied  in  a  vector  field  when  the  vector  is  a  gradient. 

130.  Reduction  of  a  line  integral  around  a  closed  loop  to  a 
surface  integral  over  a  diaphragm  bounded  by  the  loop. — A  very 
important  transformation  in  the  theory  of  electricity  and  mag- 
netism is  the  transformation  of  a  line  integral  to  a  surface  integral 
or  vice  versa.  Let  E  be  a  distributed  vector,  and  let  us  consider 
its  fine  integral  around  a  closed  curve  or  loop  AB,  Fig.  115,  the 
heavy  arrow  showing  the  direction  in  which  the  line  integral  is 
taken.  Imagine  a  diaphragm  of  any  shape  whatever  stretched 
across  the  loop  AB,  and  imagine  this  diaphragm  to  be  divided 
up  into  small  meshes  as  if  the  diaphragm  were  made  of  wire 


240 


CALCULUS. 


gauze.  Then  the  integral  of  R  round  the  loop  AB  is  equal  to 
the  sum  of  the  integrals  of  R  round  the  individuul  meshes  of  the 
diaphragm,  the  integrals  round  the  meshes  being   taken   in  the 

same  direction  as  the  integral  round 
the  loop  as  indicated  by  the  small 
curled  arrows.  This  proposition  is 
evident  when  we  consider  that  every 
line  dividing  two  meshes  is  inte- 
grated over  twice  in  opposite  direc- 
tions (see  Art.  123),  and  when  we 
consider  that  in  integrating  round 
the  various  meshes  we  eventually 
integrate  along  every  portion  of  the 
loop   AB   once  in  the  direction  of  the  heavy  arrow  in  Fig.  115. 

Let  dL  be  the  line  integral  of  R  round  one  of  the  meshes, 
and  let  jR  cos  e  -  ds  be  the  Hne  integral  of  R  roujid  the  loop 
AB,    Then  from  the  above  proposition  we  have: 


Fig.  115. 


J  R  cos  €  '  ds  =  f  dL 


(1) 


But  dL  =  C  cos  €  '  dS,   according  to  equation  (1)  of  Art.  129,  so      | 
that  equation  (1)  becomes:  ■ 


J  R  cos  e  '  ds  =  J  C  cos  e  •  dS 


(2) 


This  equation  may  be  expressed  in  words  as  follows:  The  integral 
of  a  distributed  vector  R  round  a  closed  loop  is  eqvxil  to  the  integral 
of  the  curl  of  R  over  any  diaphragm  stretched  across  the  loop.  The 
first  member  of  (2)  is  of  course  a  line  integral  and  the  second 
member  is  a  surface  integral. 

The  divergence  of  the  curl  of  a  distributed  vector  is  always  and 
everjnvhere  equal  to  zero. — Consider  two  diaphragms  stretched 
across  a  closed  loop.  The  integral  of  C  is  the  same  over  both  of 
these  diaphragms  because  each  surface  integral  is  equal  to  the 
line  integral  of  R  round  the  loop.     If  the  direction  of  the  normal 


VECTOR  ANALYSIS.  241 

be  reversed  in  the  integral  over  one  of  the  diaphragms  the  integral 
will  be  reversed  in  sign  (see  Art.  125).  Therefore  the  surface 
integral  of  C  over  a  closed  surface,  namely,  the  two  diaphragms, 
with  normal  directed  outwards,  is  equal  to  zero.  That  is  C  is  a 
vector  whose  flux  out  of  or  into  a  closed  surface  is  always  equal  to 
zero.  Therefore  C  is  a  solenoidal  vector  and  its  divergence  is 
everywhere  zero.    See  Art.  127. 

131.  Vector  potential. — Let  us  imagine  a  distributed  vector  of 
which  a  given  distributed  vector  is  the  curl.  Then  the  imagined 
distributed  vector  is  called  the  vector  potential  of  the  given  distri- 
buted vector.  The  divergence  of  a  curl  is  always  and  everywhere 
necessarily  equal  to  zero,  as  explained  in  the  previous  article. 
Therefore,  to  have  a  vector  potential,  a  given  distributed  vector 

r.  .  dX   ,   dY   ,    dZ  .   _^,      ,. 
must  have  no  divergence.    But  ^ — \-  -^ — \-  -f~  is  the  divergence 

of  a  distributed  vector  whose  components  at  a  point  are  X,  Y 
and  Z,  according  to  equation  (3)  of  Art.  126.  Therefore  the 
existence  of  a  vector  potential  (of  a  given  distributed  vector)  is 
determined  by  the  condition; 

dx^  dy^  dz       "  ^  ^ 

where  X,  Y  and  Z  are  the  components  at  a  point  of  the  given 
distributed  vector. 

Example. — In  certain  cases  of  fluid  motion  each  particle  of  the 
fluid  is  rotating  at  a  definite  speed  about  a  definite  axis.  This 
spinning  motion  of  the  particles  of  a  fluid  is  a  distributed  vector* 
and  it  is  called  vortex  motion.  Now  it  was  shown  in  Art.  129, 
that  the  curl  of  the  velocity  of  the  water  at  a  point  p  in  a  rotating 
bowl  is  equal  to  two  times  the  spin  velocity  w  of  the  particle  of 
water  at  p.     That  is,  ignoring  the  factor  2,  the  curl  of  fluid  velocity 

*  Spin  is  a  vector  quantity,  its  vector  direction  being  along  the  axis  of 
spin  in  the  direction  in  which  a  right-handed  screw  would  travel  if  turned  in 
the  direction  of  the  spin. 
17 


242  CALCULUS. 

IS  the  vortex  motion;  therefore  the  vortex  motion  of  a  fluid  is  a 
distributed  vector  whose  vector  potential  is  the  fluid  velocity. 
In  this  case  the  vector  potential  (of  vortex  motion)  is  physically 
existent;  in  general,  however,  the  vector  potential  of  a  given 
distributed  vector  is  merely  imagined  to  exist.  The  idea  of  vector 
potential  is  useful  because  it  is  sometimes  helpful  to  think  of  a 
given  distributed  vector  as  a  curl. 

132.  Rotational  and  irrotational  vector  fields.  Vector  potential 
and  scalar  potential. — ^When  the  velocity  of  a  moving  fluid  has 
curl  the  particles  of  the  fluid  are  in  rotation.  In  consequence  of 
this  fact  any  vector  field  which  has  curl  is  called  a  rotational  vector  ( ^ 
field.  A  vector  field  which  has  no  curl  is  called  an  irrotational 
vector  fiM. 

In  an  irrotational  vector  field  equations  (3),  (4)  and  (5)  of 
Art.  129  reduce  to  equations  (3),  (4)  and  (5)  of  Art.  124.  There- 
fore an  irrotational  vector  field  can  be  thought  of  as  a  gradient, 
or,  in  other  words,  an  irrotational  vector  field  has  a  potential  (a 
scalar  potential). 

In  a  rotational  vector  field  equations  (3),  (4)  and  (5)  of  Art.  124 
are  not  satisfied,  and  therefore  a  rotational  vector  field  cannot  be 
thought  of  as  a  gradient,  or,  in  other  words,  a  rotational  vector 
field  has  no  potential  (no  scalar  potential). 

A  vector  field  which  has  no  divergence  is  called  a  solenoidal  vector 
fiM,  and  a  solenoidal  vector  field  can  be  thought  of  as  a  curl,  or, 
in  other  words,  a  solenoidal  vector  field  has  a  vector  potential. 

A  vector  field  which  has  divergence  is  called  a  non-solenoidal 
vector  field,  and  such  a  vector  field  cannot  be  thought  of  as  a 
curl,  or,  in  other  words,  a  non-solenoidal  vector  field  has  no  vector 
potential. 

PROBLEMS. 

1 .  A  viscous  liquid  moves  over  the  xy  plane  so  that  X  =  az. 
Find  the  curl.    Ans:  —  a. 

2.  Find  the  vector  potential  of  fluid  velocity  when  the  fluid 
is  everywhere  moving  in  the  direction  of  the  2-axis  at  a  velocity 


VECTOR  ANALYSIS.  243 

of  +  100  centimeters  per  second.  Ans.  IOO2!  centimeters  squared 
per  second  squared. 

Note. — A  distributed  vector  which  has  neither  divergence  nor  curl  in  a 
given  region  may  have  a  scalar  potential  and  it  may  have  a  vector  potential. 

133.  Character  of  a  very  small  portion  of  any  vector  field. — 

Let  X,  Y  and  Z  be  the  components  of  any  distributed  vector  v 
at  a  point  p  whose  coordinates  are  x,  y  and  z.  Then  X,  Y 
and  Z  are  functions  of  x,  y  and  2,  and  indeed  we  will  assume 
them  to  be  continuous  functions.  If  one  is  to  reach  a  clear 
understanding  of  divergence  and  curl  it  is  necessary  to  examine 
carefully  into  the  manner  of  distribution  of  v  in  a  very  small 
region  near  a  chosen  point.  It  is  most  convenient  to  locate  the 
origin  of  coordinates  at  the  chosen  point.  Then  the  coordinates 
X,  y  and  z  of  any  point  in  the  very  small  region  are  infinitesimals, 
and  the  squares  and  products  and  higher  powers  of  x,  y  and  z 
are  negligible. 

Now  the  components  X,  Y  and  Z  being  continuous  functions 
of  X,  y  and  z  can  be  expanded  by  Maclaurin's  theorem  as  ex- 
plained in  Art.  90,  and  all  terms  in  these  expansions  which  contain 
squares  and  products  and  higher  powers  of  x,  y  and  z  may  be 
discarded,  giving: 

X  =  Xo  +  aix  +  a2y  +  a^z  (1) 

F  =  Fo  +  61X  +  h,y  +  hz  (2) 

and 

Z  =  Zo  +  CiX  +  C22/  +  Cziz  (3) 

in  which  the  constant  coefficients  are  the  values  of  the  derivatives 
at  the  origin,  as  shown  in  the  following  schedule : 


(I) 


dX 
dx 

dX 

"'-dy 

dX 

"'-  dz 

^'-  dx 

^'  =  dy 

dY 
^'=  dz 

dZ 

dZ 

'^-dy 

dZ 

244  CALCULUS. 

The  "  linear  "  vector  field. — The  simplest  type  of  vector  field 
is  the  homogeneous  field;  in  such  a  field  X  has  the  same  value 
throughout  the  field,  Y  has  the  same  value  throughout  the  field, 
and  Z  has  the  same  value  throughout  the  field.  When  X,  Y 
and  Z  are  linear  functions  of  x,  y  and  z  we  have  what  is  called 
a  linear  vector  field.  Thus  equations  (1),  (2)  and  (3)  represent 
a  linear  vector  field. 

Resolution  of  a  linear  vector  field  into  simple  parts. — The  dis- 
cussion of  divergence  and  curl  in  Arts.  126  and  129  is  not  rigorous, 
and,  as  is  always  the  case  in  a  merely  plausible  discussion,  the 
harder  one  tries  to  understand  it  the  more  vague  and  unintelligible 
it  becomes.  It  is  now  proposed,  therefore,  to  establish  rigorously 
the  ideas  of  divergence  and  curl  by  considering  a  linear  vector 
field  as  expressed  by  equations  (1),  (2)  and  (3).  For  this  purpose 
it  is  necessary  to  resolve  the  given  linear  vector  field  into  three 
parts,  and  to  be  able  to  speak  of  these  parts  intelligibly  let  us 
think  of  the  given  linear  vector  field  as  the  velocity  t;  of  a  moving 
fluid,  the  components  of  v  being  X,  Y  and  Z  as  given  in  equa- 
tions (1),  (2)  and  (3).  The  three  parts  into  which  v  is  to  be 
resolved  are: 

1.  A  uniform  translatory  motion  of  the  entire  body  of  fluid. 
This  part  of  v  will  be  represented  by  Vo  and  its  components  are 
Xo,  Yq  and  Zq. 

2.  A  simple  motion  of  rotation  of  the  entire  body  of  fluid. 
This  part  of  v  will  be  represented  by  v'  and  its  components  will 
be  represented  by  X',  Y'  and  Z', 

3.  A  continuous  stretching  of  the  entire  body  of  fluid  in  three 
mutually  perpendicular  directions.  This  part  of  v  will  be  repre- 
sented by  v"  and  its  components  will  be  represented  by  X", 
Y"  and  Z", 

Discussion  of  v'. — Consider  a  rotating  body  of  which  the  axis 
of  rotation  passes  through  the  origin  of  coordinates,  and  let  cox, 
03  y  and  0)2  be  the  components  of  the  angular  velocity  of  the 
body  around  the    x,  y    and    z    axes  of  reference,  respectively. 


VECTOR  ANALYSIS. 


245 


Consider  a  point    p    of  the  body  of  which  the  coordinates  are 
X,  y  and  z  as  indicated  in  Figs.  116,  117  and  118.    The  velocity 


Fig.  116. 


Fig.  117. 


Fig.  118. 


of  p  due  to  CO  a;  consists  of  two  components  yo)x  and  —  zccx  as 
indicated  in  Fig.  116,  and  similar  statements  may  be  made  con- 
cerning Figs.  117  and  118.  Therefore,  picking  out  the  x-compo- 
nents  of  the  velocity  of  p  in  Figs.  117  and  118,  we  get  the  value 
of  X',  and  in  a  similar  manner  we  get  expressions  for  Y'  and  Z' 
as  follows: 

X'  =  "  '  —  oi^y  +  ojyZ  (4) 

y    =    +   0)eX   •  •  •    —    OiixZ  (5) 


Z'    =    —    03yX  +    £0x2/ 


(6) 


The  values  of  cox,  coj,  and  co^,  expressed  in  terms  of  the  coefficients 
in  equations  (1),  (2)  and  (3),  can  best  be  determined  after  the 
expressions  for  v"  have  been  formulated. 

Discussion  of  v''. — As  in  case  of  v'  it  is  necessary  to  find  the 
forms  of  expressions  which  give  X'',  F''  and  Z'\  For  this 
purpose  consider  three  mutually  perpendicular  axes  of  reference 
Xi,  yi  and  Zi.  Then  a  continuous  stretch  of  the  portion  of  fluid 
parallel  to  the  Xi-axis  is  represented  by  the  equation  Xi  =  axi, 
where  Xi  is  a  velocity  parallel  to  the  rci-axis  and  a  is  a  constant. 
Analogous  expressions  using  j8  and  y  as  constants  represent 
continuous  stretches  parallel  to  the   2/1   and   Zi   axes.     Therefore 


246  CALCULUS. 

continuous  stretches  in  three  mutually  perpendicular  directions 
(which  directions  are  for  the  moment  chosen  as  the  axes  of  refer- 
ence) are  expressed  by  the  equations: 

Xi  =  axi  (7) 

1^1  =  Pyi  (8) 

and 

Zi  =  721  (9) 

It  is  desired  to  transform  these  equations  so  as  to  express 
exactly  the  same  fluid  motion,  but  to  express  it  by  giving  the  com- 
ponents of  v"  parallel  to  new  axes  a;,  y  and  z,  and  as  functions 
of  the  new  coordinates  x,y  and  z.  This  transformation  is  enorm- 
ously simplified  by  expressing  the  given  fluid  motion  [equations 
(7),  (8)  and  (9)]  in  terms  of  its  velocity  potential   P,    which  is  a 

distributed  scalar  whose  Xi-gradient  is  3—  =  axu  whose  2/1-gradi- 

axi 

ent  is  ^  =  Pyi  and  whose  2i-gradient  is  -p  =  yzi.     Integrating 

these  three  differential  equations  and  ignoring  the  constant*  of 
integration  we  get: 

P  =  IciXr'^  +  \W  +  \W  (10) 

Now  to  transform  equations  (7),  (8)  and  (9)  as  stated  above  we 
need  only  to  substitute  in  equation  (10)  the  following  values 
for  Xif  2/1  and  2i  [see  equations  (11),  (12)  and  (13)],  and  then 
find  the  gradients  of  P  in  the  directions  of  the  new  axes.  In 
this  way  we  get  equations  (14),  (15)  and  (16). 

Xi  =  hx  +  m\y  -\-  niZ  (11) 

2/1  =  Ux  +  rn^y  -f  n2.z  (12) 

Zi  =  I2PC  -{-  mzy  +  n^  (13) 

where  U  mi  ni,  U  W2  ^2  and  h  rriz  riz  are  the  direction  cosines  of 

*  The  three  equations  are  partial  differential  equations,  but  the  functions 
of  integration  reduce  to  a  single  imdetennined  constant. 


I 


VECTOR  ANALYSIS.  247 

the  new  x,  y  and  z  axes  referred  to  the  old  Xi,  2/1  and  Zi  axes. 

X"  =fx  +  py  +  qz  (14) 

F'  =  px  +  gy  +  rz  (15) 

Z"  =  qx  +  ry-\-hz  (16) 

The  coefficients  in  these  equations  are  of  course  expressions  involv- 
ing a,  jS  and  7  and  the  direction  cosines  in  equations  (11),  (12) 
and  (13),  but  they  are  represented  by  the  single  letters  /,  g,  h, 
p,  q  and  r  for  the  sake  of  clearness.  Anyway,  the  only  im- 
portant feature  about  equations  (14),  (15),  and  (16)  is  the  sym- 
metry of  the  coefficien  s  which  means  that  X'\  Y"  and  Z"  are  the 
components  of  a  fluid  motion  consisting  of  three  continuous 
mutually  perpendicular  stretches,  because  this  degree  of  symmetry 
is  the  only  necessary  consequence  of  equations  (7),  (8)  and  (9). 

Determination  of  coefficients  in  equations  (4),  (5)  and  (6)  and 
in  equations  (14),  (15)  and  (16)  in  terms  of  the  coefficients  in 
equations  (1),  (2)  and  (3). — Equations  (4),  (5)  and  (6)  and  (14), 
(15)  and  (16)  show  only  the  degree  of  symmetry  that  the  co- 
efficients must  have  in  order  that  equations  (4),  (5)  and  (6)  may 
represent  a  simple  rotation  and  in  order  that  equations  (14), 
(15)  and  (16)  may  represent  three  continuous  mutually  perpendicu- 
lar stretches,  and  the  coefficients  in  equations  (4),  (5)  and  (6) 
and  (14),  (15)  and  (16)  may  be  thought  of  as  undetermined. 
It  remains  to  determine  these  coefficients  cox,  co^,  co^,  /,  g,  h,  p,  q 
and  r,  so  that  Vo,  v'  and  v"  may  be  component  parts  of  the 
given  fluid  velocity  v  which  is  represented  by  equations  (1), 
(2)  and  (3).  To  do  this  add  equations  (4)  and  (14)  and  place  the 
coefficients  of  x,  y  and  z  in  the  resulting  equations  equal  to  the 
coefficients  of  x,  y  and  2,  respectively,  in  equation  (1);  and 
proceed  in  a  similar  manner  with  the  other  pairs  of  equations. 
In  this  way  we  get  the  following  schedule  of  equations: 

f  =  ai        p  —  coz  =  az        q  +  o)y  =  as^ 

P  +  03z  =  hi  g  =  ^2  r  -   COx  =  63  p  (II) 

q  —  03y  =  Ci         r  +  Wx  =  C2  h  =  C3J 


248  CALCULUS. 

from  which  we  get  the  following  values: 

w,  =  i(c2  —  hz) 
o)y  —  \{az  —  ci) 


Or,  using  the  values  of  ai,  a^,  03,  61,  62,   etc.,  from  the  schedule  I 

of  equations,  we  have: 

dZ      dY 

2-^  =  dy       dz 

(17) 

dX      dZ 

2co«  =  -3 3- 

^       dz       dx 

(18) 

_          dY      dX 
2"^  -  dx       dy 

(19) 

and  from  equations  II  we  get  also: 

dX 
■'       dx 

(20) 

dY 
^~  dy 

(21) 

dz 

(22) 

P  =  ^{dx  +  dy) 

(23) 

JdX  ,  dZ\ 
^  =  Kdz+dx) 

(24) 

,/dZ  ,  dY\ 

(25) 

PROBLEMS. 

1.  Find  the  flux  of  v"  out  of  a  rectangular  parallelopiped 
which  is  I  feet  long  in  the  direction  of  the  a:i-axis,  w  feet  wide 
in  the  direction  of  the  2/1-axis,  and  t  feet  thick  in  the  direction 
of  the  2i-axis,  the  constants  in  equations  (7),  (8)  and  (9)  being 
expressed  in  reciprocal  seconds.  Ans.  lwt(a  +  18  +  7)  cubic  feet 
per  second. 


VECTOR  ANALYSIS.  249 

2.  Find  the  divergence  of  v" ,    Ans.  a  +  /S  +  7- 

"Note. — It  is  here  intended  that  the  student  find  the  flux  $  which  comes 
out  of  a  given  volume  t,  and  then  determine  the  limit  of  -  as  t  approaches 
zero.    It  is  simplest  to  take  for  r  the  rectangular  parallelopiped  of  problem  1. 

3.  Show  that  the  integral  of  v"  around  any  closed  curve  what- 
ever is  zero. 

^ote. — Choose  the  axes  of  reference  as  in  equations  (7),  (8)  and  (9).  Let 
ds  be  an  element  of  the  closed  path  or  curve.  Then  the  scalar  part  of  the 
product  v"  .  ds  can  easily  be  found,  and  it  is  easy  to  show  that  the  integral 
of  this  scalar  product  aroimd  the  closed  curve  is  zero. 

4.  It  is  evident  that  the  velocity  v\  which  is  a  simple  motion 
of  rotation,  has  zero  flux  into  or  out  of  any  closed  region  whatever. 
Therefore  the  divergence  of  v'  is  everywhere  equal  to  zero.  Find 
the  line  integral  of  v'  around  the  ellipse  of  Fig.  122,  the  radius 
of  the  cylinder  being  r,  the  axis  of  the  cylinder  being  parallel  to 
the  axis  of  rotation  of  the  fluid,  and  the  angular  velocity  of  rotation 
being   co.     Ans.  l-KV^iti. 

134.  Proof  that  ^ -j-  and   -i, -r-   are  identical  when  z  is 

dy  •  dx  dx  •  dy 

a  continuous  function  of  x  and  y. — This  proposition  may  be 

stated  so  as  to  appeal  to  one's  geometric  sense  as  follows:   Let   z 

be  a  function  of    x    and    y    and  let  this  function  be  represented 

by  a  hill  built  upon  the   xy  plane.     Then  z   is  the  height  of  this 

hill  above  the  point    p'    in  the  base  plane,    x    and    y    being  the 

coordinates  of  p'   as  shown  in  Figs.  41a  and  416  in  Art.  62.    Let 

x(^-p\  be  the  a;-component  of  the  slope  of  the  hill  at  the  point  p 

(see  Figs.  41a  and  416),  and  let    ^(  =  ^)    be  the  ^/-component 

of  the  slope  of  the  hill  at  p. 

Figs.  119  and  120  represent  a  top  view  of  the  hill.  Let  dz  be 
the  difference  of  level  of  the  points   p   and   q   on  the  hill  in  Figs. 


250 


CALCULUS. 


119  and  120.    Let  us  travel  from  p  to  g  along  the  sides  a  and  h 


x^axU 


dx 


/>>. 


<^-»e 


r    X 


e  9 A  X-^^'dy 


IHveu 


^X-axis 

AS 

dx^ 

< 

y-axl 

"a 

c          b 

Y 

8 

^^gM 

Fig.  119. 


Fisc.  120. 


of  the  infinitesimal  square,  and  the  rise  will  be; 


dz  =  X  •  rfa;  +     F  + 


dY 


y    '  dx 


dx]  '  dy 


Let  us  travel  from  p  to  q  along  the  sides  c  and  e  of  the  in- 
finitesimal square,  and  the  rise  will  be: 

dz  =  Y  '  dy  ■{-  (x  -\-^  '  d^  '  dx 

Now  if  z  is  a  continuous  function  of  x  and  y  these  two  expressions 
for  dz  must  be  identical,  that  is,  one  must  rise  by  the  same  amount 
in  going  along  a  and  6  as  in  going  along  c  and  e  in  Figs.  119 


{ 


and  120.    Therefore 
dz 


dY  ^  ,  1  .      ^^ 

-J-    must  be  equal  to    -y- . 
dx  ^  dy 


But 


X  =  — 
dx 


and   7  = 


dy' 


Therefore 


^z  .,  ,  ^        d^z 

-3 -J-  must  be  equal  to  -; 

dy  '  dx  dx 


dy 


The  above  argument  is  not  entirely  rigorous  but  it  brings  out  very  cleariy 
the  geometrical  significance  of  equations  (3),  (4)  and  (5)  of  Art.  124.  To 
make  the  argument  rigorous  a  slightly  altered  point  of  view  suffices,  as  follows: 
The  gradient  of  the  hill  at  any  point  in  the  side  e  of  the  infinitesimal  square 
exceeds  the  gradient  at  the  corresponding  point  in  the  side  a  by  the  amount 
dX 


dy 


dy,   so  that  the  rise  along   e   exceeds  the  rise  along   a  by  the  amount 


VECTOR  ANALYSIS. 


251 


( 7P  •  % )  •  ^^-    Similarly  the  rise  along  side  b  exceeds  the  rise  along  side  c 

by  the  amount    (-j-  »  dx)  '  dy.    Therefore  the  rise  along  sides    c    and    e 

exceeds  the  rise  along  a  and  b  by  the  amoimt    l^ ^  J  •  dx  •  dy.    But 

the  rise  along  c  and  d  must  be  the  same  as  the  rise  along  a  and  6.    There- 


fore 


dX      dY 


-r-   must  be  equal  to  zero. 


dL 
dS 


135.  To  show  that  -^  is  the  resolved  part  of  a  vector  (the 


curl  of  R)  in  a  direction  normal  to  dS,  where  dL  is  the  line  integral 
of  R  around  the  boundary  of  a  surface  element  dS. — It  can  be 

shown  without  difficulty  that  the  two  parts  Vo  and  v"  of  the 
Hnear  vector  field  v  of  Art.  133  have  zero  Hne  integrals  around 
any  closed  curve.  Furthermore  any  distributed  vector  R  may- 
be looked  upon  as  having  a  linear  distribution  throughout  a 
very  small  region.  Therefore  to  establish  the  above  proposition 
it  is  sufficient  to  consider  the  portion  v'  of  a  linear  vector  field. 

Now  v'  is  a  simple  motion  of  rotation  about  a  definite  axis. 
Let  CO  be  the  angular  velocity  of  this  rotation.  The  line  integral 
of  v'  around  the  circle  cc  in  Fig.  121  is  equal  to  27rr^aj,   so  that 


cylinder 


Fig.  121. 


normal  to  ^-^  ^  >^ 
Fig.  122. 


the  line  integral  divided  by  the  area  of  the  circle  is  equal  to  2a>, 
as  explained  in  Art.  129.  Consider  a  cylindrical  surface  of  which 
the  end  view  is  the  circle    cc    in  Fig.  121.    A  side  view  of  this 


262  CALCULUS. 

cylinder  is  shown  in  Fig.  122.  Let  dS  be  the  area  of  an  oblique 
plane  section  of  the  cylinder.  Let  dL  be  the  Une  integral  of 
v'  around  the  boundary  of  dS.    It  can  be  shown  without  difl&culty 

that  dL  is  equal  to  2Trr^o).    But  the  area  of  dS  is ;:.    There- 
cos  6 

fore  -75  is  equal  to  2co  cos  6.  That  is,  2w  is  a  vector  whose 
component  in  the  direction  of  the  normal  to  dS  is  equal  to   -75-. 

136.  The  distortion  of  a  small  portion  a  body  which  has  been 
twisted  or  bent  in  any  manner  whatever. — Consider  a  particle  p 
of  an  unbent  beam.  Imagine  the  beam  to  be  bent  or  twisted  in 
any  manner  whatever,  and  let  1;  be  a  line  drawn  from  p  to  p', 
where  p'  is  the  position  of  the  particle  after  the  beam  is  bent. 
Then  the  line  v  is  a  vector,  it  is  called  the  displacement  vector  of 
the  particle  p,  and  it  may  be  thought  of  as  being  at  p  because 
it  refers  to  the  point  p.  Now  v  has  a  definite  value  and  a  definite 
direction  at  each  point  of  the  unbent  beam,  therefore  it  is  a  distri- 
buted vector.  That  is  the  unbent  beam  may  be  thought  of  as  a 
vector  field  in  its  relation  to  the  bent  or  twisted  beam. 

Consider  a  very  small  portion  of  the  unbent  beam.  The 
displacement  vector  v  may  be  thought  of  as  having  a  linear  distri- 
bution throughout  this  small  portion  of  the  beam  as  explained 
in  Art.  133.  Therefore  the  displacement-vector  field  v  may  be 
resolved  into  three  parts  as  explained  in  Art.  133.  One  part  Vq 
is  a  simple  translatory  displacement  of  the  entire  small  portion 
of  the  beam;  another  part  v'  is  a  simple  rotation  of  the  small 
portion  of  the  beam  about  a  definite  axis;  and  a  third  part  v" 
consists  of  three  mutually  perpendicular  stretches.*  This  is  a 
fundamental  proposition  in  the  theory  of  elasticity.! 

*  The  expression  cordinuom  stretches  as  used  in  Art.  133  referred  to  what 
takes  place  in  a  rubber  band  while  it  is  being  stretched,  and  the  word  stretch 
as  here  used  refers  to  what  has  taken  place  in  a  rubber  band  after  it  has  been 
stretched. 

t  See  the  chapter  on  elasticity  in  Franklin  and  MacNutt's  Mechanics  and 
Heat,  The  Macmillan  Co.,  New  York,  1910. 


I 
I 


VECTOR  ANALYSIS.  253 

PROBLEM. 

1.  One  end  of  a  rubber  band  is  fixed  and  the  band  is  stretched 
by  moving  the  other  end  of  the  band.  The  initial  length  of  the 
band  is  10  inches,  the  increase  of  length  is  0.5  inch.  Express  the 
displacement  of  each  particle  of  the  stretched  band  in  terms  of 
the  initial  distance  x  of  the  particle  from  the  fixed  end  of  the 
band,  ignoring  the  lateral  contraction  of  the  band.  Ans.  Displace- 
ment =  O.OSic. 


APPENDIX  A. 
PROBLEMS  GROUP  1. 

(After  Art.  34.) 
Differentiate  the  following: 

1.  2/  = 


n+  1' 


dx 


g  =  20.3(.  +  l). 


3.  y  =  6a;8  -  5x^  +  4^^  -  Sx^ 
dy 


dx 


4,y  =  x^x  +  7), 

5.  t/  =  3a;'(a;2  -  5a;), 

6.  e  =  ar*(6r3  -  cr  +  d),  7^b<*-^ 

^  =  ar3(76r3  -  5cr  +  4d). 
ar 

7.  0  =  (r  +  3)(2r  +  5), 

^^       ^     I   11 
-=-  =  4r  +  11. 

ar 


8.  e  =  (2r2  +  3r)(3r2  +  2r), 


^  =  3r(8r2  +  13r  +  4). 


CALCULUS. 


dr 


^  =  (r2  -  r  +  l)(5r2  -  3r  +  1). 
ar 


37  =  2act  -{-  ae  +  be, 
at 


9.  0  =  (2  -  r)(3  -  r), 

dr  '' 

10.  0  =  (r^  -  r2  +  r)(r2  -  r  +  1), 

de  ^ 

dr 

11.  s  =  iat  +  h)(ct-\-e), 

ds  ^ 
dt 

12.  «  =  (a  -  60(c  -  eO, 

ds  _ 
dt  ~ 

13.  s  =  (3^  -  2<  -  1)(1  -  3^2  4-  5^3)^ 

14.  8  =  (2  H-  0^ 

15.  5  =  (1  +  2x)^ 

16.  8  =  ^(2<  4-  3)«, 


5?  =26e<-  (ae  +  bc). 
at 


5^  =  75<*  -  76^3  +  3^  +  12^  -  2. 
at 


^  =  3(2  +  t)\ 


^  =  6(1  +  2x)\ 
ax 


I  =2«(7<  +  3)(2<  +  3)^ 


17.  s  =  a«H6^  +  c«  +  e), 


^  =  a^(56^  +  4d  +  3e). 


18.  2/  =  (1+  2x)H2x2  +  1)\ 


^  =  2(1  +  2a;)2(2x2  +  1)3(3  +  Sx  +  22x2). 
ax 


19.  2/  =  (ax  +  by{a  +  6x)3, 


^  =  (ax  +  6)  (a  +  6x)2(5a6x  +  2a2  +  36^). 
ax 


APPENDIX  A 

L. 

2°-^  =  x  +  2' 

A^: 

dy             2 

v- 

da;       {X  +  2Y' 

^t               CI  —  X 

dy        a  —  h 
dx~  (b-xy 

^^-  ^"  (2x  +  5)4' 

dy           2(a;  +  3)(2a;  +  7) 

dx                  {2x 

+  5)» 

23    .-(^^  +  ^)' 

t 

z'  1  +  a;  \ 


d^  _  {ax  +  bYiSae  -  76c  -  4aca;) 
dx  ~  (ex  +  ey 


25. 


26. 


/2x-  1\ 


/ax^  -hx  +  cY 


dy  _ 
dx 

3(1  +  xy 

{2x  +  3)* 

• 

dy 

12(2x  - 

-D' 

dx- 

(X- 

2)'    • 

dy  __  5{ax^  —  c)(ax'^  —  hx  -\-  c) 
dx  e^x^ 

27.  Find  the  slope  of  the  curve  y  =  x^  —  2x^  +  Z  at  (a) 
x  =  0,   (6)  X  =  2,  and  (c)  x  =  -  2.     Ans.  (a)  0,  (6)  4,  (c)  -  20. 

28.  At  what  angles  does  the  curve  y  =  x{x  —  2){x  —  3)  cut 
the  a;-axis?  Ans.  (a)  At  a;  =  0,  slope  is  6,  (6)  at  x  =  2,  slope  is 
—  2,  (c)  at  a;  =  3,  slope  is  3. 

18 


4  CALCULUS. 

2a;  4-  3 

29.  Is  -. — r-^  an  increasing  or  a  decreasing  function  of    x? 

4x  +  o 

Ans.  Decreasing. 

30.  Is  o    _t  A   ^^  increasing  or  a  decreasing  function  of    «? 

oX  -p  4 

Ans.  Increasing. 

31.  When  y  =  2x^  —  Qx  +  5,  for  what  values  of  x  is  t/  (o) 
an  increasing  function,  and  (6)  a  decreasing  function?  Ans.  (a) 
All  values  of  x  less  than  —  1  or  greater  than  +  1,  (6)  all  values  of 
X  between  —  1  and  +  1. 

PROBLEMS  GROUP  2. 

(After  Art.  89.) 

Differentiate  the  following  functions 

Note. — The  forms  referred  to  in  the  answers  to  the  following 
problems  are  found  in  the  table  of  integrals,  see  Appendix  B.    Thus 
the  answer  to  problem  4  is  given  by  form  45,  as 
dy  1 


dx 

V^ 

+  hx 

the  answer  to  problem  5  is 

found 

to  be 

dy  _ 

x 

=,  etc. 

l.V  =  ^-3x  +  l-l„ 

dy  _ 
dx 

2x-3-%-^. 

2.  J/  =  3V?-2V?, 

dy 
dx 

2Vx'          5 

dy  _ 
dx 

3      ,  ,   20     , 

b 

Form  45. 

S.  y=  Va;2  +  a^, 

Form  61. 

6.  2/  =  -  Va^  —  x^y 

Form  53. 

APPENDIX  A.                                             5 

7.  e  =  (r^-  a^y^,  ^  =  5r%r'  -  a')K 

8.  y  =  ib  V(aT6i)^  Form  42. 

9.  y  =  i  V(rc2  +  a2)3,  Form  58. 

10.  2/  =  -  J  V(a2  -  a;2)3,  Form  49, 

11.  1/  =  (a.  -  x.)t,  ^  = ^. 

d^       r3(7  -  8r) 


12.  d  ==  r^-ylr  -  r^, 


rfr        2  -^7-  _  7.2 


2(2a  -  hx)  ^a  +  hx  ^ 

14.  2/  =  — 25^ ,  Form  46. 

_            2(2a  -  Zhx)  V(a  +  M^  ^ 

15.  2/  =  -^ ^^ '-y  Form  43. 

2(8a2  -  4:ahx  +  36V)  Va  +  6rc 

16.  2/  =  ~- —         ^553 >  Form  47. 

17.  2/  =  -^ 1Q553  ^>  Form  44. 

18.  2/  =  7,/     ■   I.  x>  Form  34. 

o{a  +  oaj) 

19   ^  =  _  ^^l-~^\  Form  64. 

20.  y  --=    ,   ,1 ,  Form  66. 

aNa;2  +  a2' 

22.  „  =  -  ^'^^  -  ^,  Form  73. 
^                  ax 

-  (3^'  -  Q')^  ^  =  (a;^  -  a^)K7a;^  +  9a') 

^^'  ^  ~         a;5       '  dx       .              6a;§ 

|a2"-  aj2  <^^                  -  4a2a; 


"    NoH^T^'  S  ^  3(a2  -  a;2)§(a2"+^t 


CALCULUS. 


25.,  =  log„.n,  1  =  ^ 


26.  u  =  log  (3t;2  +  4t;3)», 


du  ^  18(1  +  2t;) 
dv        v{3  +  4t;)  ' 


27.  2/  =  ^  log  (a  +  6a;),  Form  33. 

28.  2/  =  ^  log  ^a;2  +  ^V  Form  39. 

29.  2/  =  log  (x  +  Vl  +  x^)y  Form  10. 

30.  y  =  log  (a;  +  V^~^),  Form  11. 

31.  2/  =  log  (x  +  Va;2  +  a^),  Form  60. 

32.  2/  =  -  log(^  +iJJ+l),  From  16. 

33.  2/  =  -  log  (i  +  ij^2  -  l)»  Form  15. 

35.  s  -  ti  log  (a^  -  60*     I  =  %faT-^W  "^  ^*  ^""^  ^""^  "  ^^^' 

^^•^  =  ^^^S^'  ^^^^^• 

37.  2/  =  i  log  i-±^,  Form  13. 

1         X 

38.  2/  =  -  log  — 1 — P==,  Form  63. 

39.  2/  =  -  log  "^^'  +  ^'  -  ^  Form  63. 
"a                 X 

__J___,      2cx  +  6  -  Ate2  _  ^ac  ^         .. 

40.  2/  =     h,       .     log  X r-i—F — ■  f  Form  41. 

^       V62  -  4ac    *^  2cx  +  6  +  a/^^  -  4ac 

41.  2/  =  p    a  +  6x  —  a  log  (a  +  6a;)   ,  Form  35. 

42.  2/  =  J  [log  (a  +  6a;)  +  ^^^],  Form  36. 


APPENDIX  A.  7 

43.  2/  =  ^a^  -  x^  —  a  log  — ' ,  Form  51. 

X 

44.  1/  -  I  -VSM^  +  |-  log  (a;  +   Va;2  +  a^),  Form  57. 

45.  2/  =  I  Va;2  +  a^  -  ^  log  (a;  +  V^^  +  ^2)^  Form  62. 

46.  2/  =  |(2a;2  +  Sa^)  Va:^  +  a^  +  ~  log  {x  +  Vx^  +  a^),  Form  65. 

47.  y  =  |(2a;2  +  a^)  ^2  _|_  ^2  _  |'  i^g  (^^  ^^2-^:^2)^         Form  59. 

48.  2/  =  log.  (30. +  2),  J  =  3^-2 

49.  ^  =     ^  %  =  _  L±J2g^ 

a;  log  x*  dx  {x  log  xY ' 

50.  2/  =  a:i°«=",  1^  =  loga-x*°««-i 

51.  2/  =  a^,  ^  =  2a;a-Moga. 

^"^^  ^      "^     »  da:  ~         ic         • 

53.  ^  =  c=^e^  3-  =  c^e^(l  +  log  c). 

54.  2/  =  e^'x^  -^  =  e=^a;^-K^  +  ^)' 

55.  2/  =  ^^x\  ^  =  4-a:3(4  +  x  log  4). 

_  ?!±?  dy       /-        2\   ?!±2 

56.2/  =  e«,  ^=(^l-_je-^. 

58„-5£_+3^^  #      24X-W 

5'- ^=(^''-2)'.       ,  %-W^r 


8  CALCULUS. 

11  1  dy       1  +  log  X 

60.  y  =  loglog^  -  j^,  ^  =  -^(bi^- 

61.  y  =  ^{e^  -  e-2-),  ^  =  c{e'^^  +  6-2-). 

62.  s  =  (2e2*  -  1)(1  -  3e-30,        ^^  =  4e2<  +  Ge"*  -  Qe-^'. 
/:o    -..      i^«     log  ^  ^2/ I 


MO, 

y 

-^^^l  +  logx' 

64. 

y 

=   glog„x  _   a;log««^ 

65. 

y 

e'  +  x- 

dx      a;  log  X  (1  +  log  x) 

^=  loga_e/    ,,^,_^,,^A 

dx  iC     \                          / 

^  _  2e'xf'-\a  —  x) 

dx  "  (e*  -  x")2 

66.  2/  =  x[(log  xY  -  2  log  X  +  2],  ^  =  (log  x)2. 

2/  -  log  gnx  _^  1  _^  g-nx  »  ^a-  -   g2«x  +  1  +  e'^"*  * 

riogi(e-  +  l)  __  log  (6"- -1)1 

=  log  (e"''  -  1)        %^         L     e"^-  1 e"^  +  1     J 

^^*  ^      log  (€"-  +1)'       da;  [log  (e--  +  1)]2 

PROBLEMS  GROUP  3. 
{After  Art.  42.) 
Differentiate  the  following: 

1.  2/  =  sin  hx, 

2.  2/  =  cos  nXy 

3.  2/  =  sin  3a;2, 

4.  2/  =  2  +  4  Sin  2a;, 

a;      1 

5.  2/  =  2  ""  4  sin  2a;, 


d2/ 

5  cos  5a;. 

dx~ 

—  n  sin  na;. 

dy_ 
dx 

6a;  cos  Sa;^. 

Form  21. 

Form  20. 

APPENDIX  A.  y 

dii 
6.  y  =  sin  Zx  cos  2x  +  cos  3a;  sin  2x,    3^  =  5  cos  5x. 

sin  (m  -  n)x       sin  (m  +  n)a; 
^-  ^         2(m  -  n)     ^     2(m  +  n)     '  '''' 

^  «m  (m  -  n)x  _  sin  (m  +  n)a;  ^^^^  24. 

^         2(m  -  n)  2(m  +  n)    ' 

^  cos  (m  -  n)x      cos(m  +  n)a:  ^^^^  25. 

^  2(m  -  ri)     ^      2(m  +  n)     ' 

10.  y  =  log  sin  x, 

11.  2/  =  —  log  cos  a;, 

12.  s  =  sin(<2  -st  +  2), 

13.  s  =  log  cos  (1  -  f), 

14.  y  =log(a  sin2  x  dy  ^  2(a  —  h)  tan  x 

+  &  cos^  x),  dx        a  tan^  x  -{-  h   ' 

15.  w  =  2x2  sin  2a;  dy       .  ^        ^ 

,  o         o         .    ^     /  =  4a:2  cos  2a;. 
+  2a;  cos  2a;  —  sm  2a;,  dx 

16.  2/  =  6=^  sin  (2  -  a;2),       ^  =  e^  [sin  (2-a;2)  -2a;  cos  (2-a;2)]. 

^o        re        -«^      ^^       0[2  cos  (e«  -  6"^)  -  ^e^  +  6-«) 

17.  r  -  02  cos  (e^  —  e "),    t^  =        •    /  «       _flM 

^  ^'    dd  sin  (e^  —  e  '')]. 

,  cos  a;  dy  sin  c 

18.  i/  =  log 


Form  8. 

Form  7. 

J^  =  (2^  -  3)cos(i2  _ 

3«  +  2). 

1  =  2aan  (1  -  i2). 

cos  (a;  +  c)  '  da;  cos  x  cos  (a;  +  c) 

19.  V  =  sin'  2x  cos^  3a;,  i^  =  6  sin^  2a;  cos  3a;  cos  5a; . 
^  aa; 

20.  v  =  e^^'^^^  -^  =  2a;cosa;2.e^^^^ 
^  dx 

*  ^         ^  sin  a;  +  cos  a; '  da;  cos  2a;  * 

,      sin  J(a;  —  a)  d?/  sin  a 

22    1/  =  log — -  — ^  =  

'  ^  sin  |(a;  +  a)'  dx  2  sin  J(a;  —  a)  sin  |(a;  +  a) 


10 


CALCULUS. 


23.  The  lengths  of  crank  radius  and  connecting  rod  of  a  steam 
engine  are  2  feet  and  10  feet  respectively,  as  shown  in  Fig.  p23, 
and  the  crank  revolves  at  the  uniform  speed  of  2  revolutions  per 
second.  Find  the  velocity  of  the  piston  for  the  following  values 
of  the  angle  6.  (a)  6  =  0,  (h)  6  =  45°,  (c)  6  =  90°  and  (d)  6  = 
135°.  Ans.  (a)  zero,  (h)  20.32  feet  per  second,  (c)  25.13  feet  per 
second,  (d)  15.24  feet  per  second.     Fig.  p2Z  is  1  J"  high  X  4"  long. 


Fia.  p23. 

PROBLEMS  GROUP  4. 
{After  Art.  43.) 
Differentiate  the  following: 


1.  2/  =  cot  X, 

2.  y  =  sec  x, 

3.  2/  =  cosec  Xf 

4.  2/  =  vers  x, 

5.  2/  =  tan  {x^  —  2a:), 

6.  2/  =  sec"  nx, 

7.  2/  =  log  tan-, 


J    =  —  cosec^  X. 
ax 

dy  . 

-J-  =  sec  X  tan  x, 

ax 

dy  . 

:r  —  —  cosec  x  cot ». 
aa; 


=  smx. 


dy 
dx 

^  =  2(x  -  1)  sec2  (a;2  -  2x). 

dy        „ 

■f^  =  rv-  sec"  na;  tan  nx. 

ox 


Form  la 


APPENDIX  A. 


11 


S,  y  =  log  tan  ^  + I j, 

9.  y  =  log  (sec  x  +  tan  x), 

10.  y  =  log  (cosec  x  —  cot  x), 

11.  2/  =  e^*  cosec  2x, 


Form  18. 

Form  18. 
Form  19. 


4y  _ 


12.  1/  =  (tan w— Scot w)  Vtan^.-^p  =  r— — v— . 

'  dw      2  tan5  w 


da; 
dy 


=  2e2'  cosec  2x(l  -  cot  2x), 


13.  s  =  <"  cosec"*  nt, 
2taiid 

15.  1/  =  log  [tan  (e^  +  S)*]^, 


16.  e  =  log 


tan  ^  —  2 


17.  r  = 


2  tan  ^  -  1 
a  sin  ^  +  6  vers  ^ 


-T.  =ni"~^  cosec*"  n<(l  —  mi  cosnQ. 

di/  _      2  sec^  e 
Te~{l  -  tan2  ^)2* 

dx     sin  [2(e^  +  3)i]* 

dg^         3 

dr      4  —  5  sin  r 


dr 


2ab  vers  6 


a  sin  0  —  6  vers  0  d^      (a  sin  0  —  6  vers  ^)2' 


PROBLEMS  GROUP  5. 
(After  Art.  U-) 
Differentiate  the  following: 

dy  1 


1.1/  =  COS~^  X, 

2.  y  =  tan~^  a;, 

3.  2/  =  cot~^  x, 

4.  y  =  sec~^  a;, 

5.  2/  =  cosec"^  a;, 

6.  2/  =  vers~^  a;, 


da; 


Vl-  a;2 


Form  12. 

dy 
dx 

1 

l^x^' 

Form  14. 

dy  _ 

1 

dx 

x^lx''  - 

1 

Form  17. 


12  CALCULUS. 


X 

7.  2/  =  sin-i  -,  Form  52. 

8.  y  =  -tan-i-,  Form  37. 

c  c 


2  ,  2cx  +  h 


Form  40. 


1  X 

10.  J/  =  -  sec-i-,  Form  67. 

o  a 

11.  y  =  vers-i  -,  Form  70. 


12.  2/  =  x sin-i  X  +  Vl  -  a^,  Form  27. 

13.  2/  =  x  cos-i  a:  —  V 1  —  x^  Form  28. 

14.  2/  =  I  Va2  -  x2  +  ^sin-i-,  Form  48. 

15.  2/  =  -  I  Vo^"^^  +  ^sin-i -,  Form  54. 

16.  2/  =  1(2x2  -  a^)  Va2-x2  +  |^sin-i  -,  Form  50. 

17.  2/  =  J(5a2  -  2a;2)  Va^-rc^  +  ^'  sin-^-,  Form  55. 

o  o  O 

18.  to  =  COS"*  Vvers  t  -jj=  —  h  Vl  +  sec  i. 

^       ,3^  -  2   ,       ^  ,  3(9  -  12        dr       ^ 

19.  ,  =  tan-^-  +  cot-  -^^^,       ^  =  0. 


20.  r  =  cos-i-3-  -  2^-^,  rf",  =  aJ^  -  r 

21.  s  =  t^  sec-i  I  -  2  Vi2-4,  ^=2^  sec-i|. 

22.  0)  =   V2[sec-*  (sec  i  +  tan  01, 


dt       V  (sec  t  +  tan  0  tan  t 
«.           ^      ,4  +  5  tan  x  dy  S 

23.  2/  =  tan    3 d^  =  5  +  4sin2x 

OA  -  -l!i±£l'  ^  -         -  2 

Z4.  s  -  tan    ^,  _  ^_,,  ^^  -  ^2,  ^  g-2f 


APPENDIX  A 

25.  y  =  tan-'  (|tan  x)  -  tan-'^,     g  =  - 

26.  y  =  tan 


13 


ab 


sin^  X  +  ¥  cos^  x' 
J  g^  tan  a;  -  6^  dy  a6 


a?)(l  +  tan  xY 


27.  1/  =   V2ax  —  x^  +  a 


vers" 


28.  2/  =  —  V  2aa;  —  x^+  a 


vers"*  — . 


X  —  a    ITT 5  ,  a^  ,  X 

29.  2/  =  — ^ —  V  2ax  —  a;^  +  —  vers~i  -, 

30,  y  = ' — ^ V  2ax  -  a;2  +  —  vers-^  -, 


dx      o?  sin^  X  ■\-\P-  cos^  x 

Form  72. 

Form  71. 

Form  68. 

Form  69. 


PROBLEMS  GROUP  6 

{Ajter  Art.  ^.) 

Differentiate  the  following: 
1.  2/  =  ^'^'j 


2. 0  =  (1  -  ty-', 

3.  s  =  Zt'\ 

4.  2/  =  x^^''^*^", 

5.  2/=  (iogxy°«- 

6.  2/  =  (3x2  +  1)2^^-2), 


^  =  nx"*(l  +  log  x). 
ax 

I  =  -  (1  -  0^-'[i  +  log(i  -  01 

|  =  3<^'ni  +  31og<]. 
^=  (n+l)(logx)«x^'"«*>"-i 

^'[n-log(logx)]. 


dy  ^  (log  x) 
dx 


dy 
dx 


(3x=  +  !)«-«' {2  log(3a;»  + 1)  +  ^3^^^-^}  • 


14  APPENDIX  A. 

PROBLEMS  GROUP  7. 
{After  rule  III  of  Art.  67.) 
Integrate  the  following  differential  equations. 

1.  dy  =  (3x'-7^+ 1  -  I'jdx,  y  =  ^-1^+  log^  +^  +  C. 

2.  dy  =  (2  -  ^)V  ■  dx,  j,=^_^  +  |'+C. 

3.  dy  =  3(2j?  +  S)3?  •  dx,         y  =  ^2?f-±-^+C. 

dx 

-     ,  dx 

^     ,  X  '  dx 

7.  dy  =  Va  +  hx'  •  dx,  Form  42. 

dx 

8.  di/  =     ■        ,   ,  Form  45. 

VaH-  hx 

9.  dy  =  X  Va2  -  a;^  •  c?x  Form  49. 

3/  *  dx 

10.  di/  =     ■  ,  Form  53. 

Sa^  —  x^ 

dx 

11.  dy  =—-==,  Form  73. 

^<2ax-x^ 

12.  dy  =  a'  '  dx,  Form    4, 

13.%=(e-  +  a-  +  36-)dx,    ,  =  ^^+^_^^+C. 

15.  dy  —  sin  X  cos  x  '  dx  y  =  — ^ —  +  C 

16.  dy  =  tan  a;  •  dx,  Form  7. 
JVote. — ^Take  tan  x  = . 

C06  X 


APPENDIX  A.  15 

17.  dy  =  cot  X  •  dXj  Form  8. 

-o     ,  .  .     ,  ,  2qo^^x      cos^x  ,   ^ 

18.  a?/  =  sm^  xdx,  y  =  — cos  x  H ^ ^ f-  C. 

iVote. — Take  sin"  x  =  iX  —  cos^  x)"  sin  x. 

,-,    J         -1         K        J  sin^a;      2siD^a;  ,  sin®  a;  ,  ^ 

19.  dy  =  sm*  a;  cos^  x  -  dXy  y  =  —z = 1 ^ h  C. 

20.  dy  =  sin^  a;  •  c^a;,  Form  20. 
1  —  cos  2a; 


iVote. — Put  sin^  x  = 


2 


21.  c?2/  =  cos^  X  '  dx,  Form  21. 
-  -     ,             fi         7                 5a;  ,  sin  2a;     sin^  2a;  ,  3  sin  4a;   ,   ^ 

22.  dy  =  cos' X'dx,      2/ =  jg  + -j is"  "*"  "~64~  "^  ^• 

23 .  (??/  =  sin  mx  sin  na;  •  da;,  Form  24. 
iVote. — Put  sin  mx  sin  nx  =  }^  cos(m  —  n)a;  —  3^  cos  (m  +  n)a;. 

- .     ,         ,     J.         ,  tan^  a;       tan^  a;   '  ,    ^ 

24.  ai/  =  tan^  x  -  dx,  y  =  — -. ^r log  cos  x  -^  C. 

4:  ^ 

Note. — Put  tan"  x  =  tan'  x  (sec^  x  —  I). 

cot^  X 

25.  dy  =  cot^  X  '  dx,  y  =  x  -\-  cot  x ^ h  C. 

o^    J  fi        ,  ^        X       2  cot^  a;     cot^  x 

26.  ai/  =  cosec^  x  -  dx,  y  =  C— cot  x i^— ^ — . 

o  5 

iVofe. — Put  cosec*  x  =  (1  +  cot^  x)''. 

ofT    J        J.     1  fi        J  tan''  x  ,  tan^  a;  ,  tan^  x  ,  ^ 

27.  dy  =  tan'  a;  sec^  x  -  dx,  y  =  — 1 1 ^ f-C 

4  o  o 

iVo^e.—Put  sec*  x  =  (1  +  tan^  a:)2. 

v,o    J        ,     .  ,        J  sec^  X     2  sec^  a;  ,  sec'  x  ,  ^ 

28.  dy  =  tan^  a;  sec'  x  -  dx,  y  =  —^ = 1 ^r — |-  C. 

7  o  o 

Note. — Put  tan^  x  =  (sec^  x  —  1)^  and  write  tan"  x  sec'  x  =  (sec*  x  —  1) 
sec^x-sec  xtanx-dx. 

29.  dy  =  -p=^=,  2/  =  i  sin-i  ~  +  C, 

^-     ,  dx  1  , 

*  "  VliP'^l'  2^  =  5  log  (5x+V25x2-4)  +  C. 


16  APPENDIX  A. 

^^' '^'' =  4T25:^'  J/ =  2  tan-'  2- +  C. 

,~    J  dx  1  ,      2  +  6a;  ,    - 

35.  .,  =  ^^  .  dx,   ,  =  I  log  (4x-6)+^-?^  '°^  £|+^- 

36.  Find  the  length  of  the  arc  of  the  parabola  in  Fig.  13  on  page 
24  from  x  =  0  to  x  =  10  inches,  the  value  of  A;  being  2  when 
X  and  y  are  both  expressed  in  inches. 

Ans.    201  square  inches. 

PROBLEMS  GROUP  8. 
(After  rule  IV  of  Art.  67.) 
Verify  the  following: 

1.  Jx  smx  •  dx  =  —  a;  cos  a;  +  sin  a;  +  C. 

Note. — Use  rule  IV.    Try  u  =  em  x,   dv  =  x-dx]  also  try  u  =  x,  dv  = 
mnx-dx. 

2.  fx  COS  a;  •  da;  =  a;  sin  a;  +  COS  a;  +  C 

/x^ 
x^logx  '  dx  =  -J  (log  X  —  I)  +  C. 

4.  fx{^'  +  e-'')dx  =  I  (e^x  _  g-sx)  _  i  (^sx  +  g-3x)  +  q. 

5.  J* log  (ax  +  6)  •  cte  =  -  (ax  +  h)  log  (ax  +  &)  —  a;  +  C. 

6.  fa^smx'dx=  —  x^ cos x  +  Sx^ sin x  +  6x cos x  —  6 sin x  +  C. 

Note. — Applying  rule  IV,  taking    u  =  x^    and  dv  =  smx'dx,    we  get  an 
expression  involving  Jx^  cos  x-dx.    This  expression  can  again  be  reduced  by 


APPENDIX  A  17 

applying  rule  IV  (to  the  part  to  be  integrated),  making  w  =  a;^  and  dv  » 
cosx'dx]  and  so  on. 

7.  fx^  sin  2x  '  dx  =  §(2a;2  -  1)  sin  2x  -  iC^s^  -  3x)cos  2x  +  C, 

„     /•  „^   .    ,        J        e°^(a  sin  6x  —  6  cos  6a;)  ,  ^ 

8.  I  e«^  sm  hx  •  (ix  =  — ^^ .   ,   ,, ^  +  C. 

•^  a^  -jr  0^ 

Note. — Integrate  by  parts,  taking  u  =  e<**,  then: 

/,  •    I.     J             e""  C08  hx   ,    a  r    ^         ,      ,  ,^ . 

e'^'^embx-dx  = ^^ ^  hj  ^     ^°^  oa;*aa;  (1) 

Integrate  by  parts,  taking  u  =  sin  bx,  then: 

/e"*  sin  bx'dx  = /  e"*  cos  bx'dx  (2) 

a  a«/ 

Take  the  sum  of  equation  (1)  multiplied  by  ¥  and  equation  (2)  multiplied 
by  a^  in  order  to  eliminate  the  term  containing  J  e°*  cos  6a:  •  dx  and  the  result 
given  above  is  obtained.  By  subtracting  equation  (2)  from  equation  (1), 
J  e"*  sin  bx'dx  can  be  eliminated  and  the  value  of    I  e"*  cos  bx-dx  obtained. 

9 .  J  e^^  cos  5x  '  dx  =  -^{5  sin  5x  +  3  cos  5x)  +  C. 

e~2*  sin  a;  •  (ia;  = ^  (2  sin  x  +  cos  ic)  +  C 

X 

11.  fe~^  COS  ^'  dx  =  -TY"  (2  sin  r-  —  3  cos  ^j  +  C. 

PROBLEMS  GROUP  9. 

(Special  methods  a  of  Art.  67.) 

Verify  the  following: 

r     5a; +  4  _    /"3  -  c?a;        /•2j^ 

^•Ja;2  +  x-2*'^~Ja;-l"*"Ja;  +  2 

=  3  log  (a;  -  1)  +  2  log  (a;  +  2)  +  C. 

Note. — A  full  discussion  of  the  integration  of  rational  fractions  is  given  in 
Byerly's  Integral  Calculus.     See  note  on  page  86. 

In  this  particular  case  the  denominator  is  equal  to    (x  —  1)  (x  +  2)    and 


18  APPENDIX  A. 

5x4-4 


««  +  a;-2 


may  be  broken  up  into  what  are  called  partial  fractions  as  follows: 
&C  +  4     _     A  B 


a:*  +  x-2      X  -  1      x-{-2 


Clear  of  fractions  and  equate  coefficients  of  like  powers  of    x    and  we  find 
A  =  3  and  5=2. 

2.  f^^^^l^-dx  =  \og(x  +  i)  +  \og(x+l)  +  C. 

+  \  log  (x  -  2)  +  C. 

=  ^'  +  15x  -  6  log  (x  -  1)  +  41  log  (x  -  2)  +  C. 

Note. — \Mienever  the  degree  of  the  numerator  of  a  rational  fraction  is 
gre^ater  than  or  equal  to  the  degree  of  the  denominator,  the  fraction  should  be 
reduced  to  a  mixed  quantity. 

5-/^^^-'i-  =  -  +  glog(--3)-|log(.  +  2) 

+  \  log  (x  +  1)  +  C. 

'^ •  /  (X  -  IRx'  +  1 )  •  "^  °  ^°g  ^^  ~  ^^  "  W^) 

-  I  log  (x^  +  1)  +  C. 

X* 

Note. —     __  ^       -      ^    is  broken  into  partial  fractions  by  putting  it  equal 

.^     A  B        ,Cx-\-D 

^  x-l'^  ix-1)*'^  x«  +  1  * 

«•  J  i(^^-:^4?  •  ^^  =  -  i^^4  +  ^^S^^^  + ^- 


APPENDIX  A.  19 

PROBLEMS  GROUP  10. 
(Special  methods  b  of  Art.  67.) 

Integrate  the  following  and  for  answers  see  specined  forms  in  the 
table  of  integrals  of  Appendix  B. 

1.  /Vo^-^^  •  dx  Form  48. 

Note. — Let  a:  =  a  sin  ^,  then  dx  —  a  cos  d-dd;  and  we  get: 

(^CL^  —  x^-dx  =    C<a*  —  a'  sin*  d-a  cos  6-6$ 
=  a^  fcoa^d'dd 


=  a^ 


X,    Ix^a^  -  X* 


» -  xn 

o»         J 


a*    .       .  X    .     1       ^r-r -r- 


2.  Jx^^a"  -  x^  '  dx, 

,      rVa2  -  x^      , 

3.  J— ^—  -da:, 

4.  f{a^  -  x^)i  •  dx, 

6.  fV^M-  a^  •  dx, 

7.  faj^  Va;2  -{- a^  •  dic, 
8. 


9. 

10 


\x^  +  a2 
dx 

dx 


/rix 
(x2-a2)r 


^           a^x^      ' 
X 


Form  50. 
Form  51. 
Form  55. 
Form  56. 
Form  57. 
Form  59. 
Form  62. 

Form  64. 
Form  66. 


19 


20  APPENDIX  A. 

PROBLEMS  GROUP  11. 
(Special  methods  c  of  Art.  67.) 
Verify  the  following: 

J      r  x^  '  dx  ^{x-\-  ay-^  {x  +  ay-^ 

*  J  (a;  +  a)"*        4  —  m         ^     3  —  m 


2  —  m  1  —  m 


i\rote. — ^Let  X  -^  a  =  y,    dx  =  dy. 
X  '  dx         3   ,     .   ,   ..       3 


2- /(STS-r|.(« + ^*)'- i  («  +  ''-)'  + ^- 


JVote. — Let  a  -\-hx  =  y,    h'dx  =  dy, 
1.    r.f"    ,  =  tan-i  e-  +  C. 

J  6^  +  e-* 

A^ofe. — Let  e*  =  y,  x  =  log  w,  dx  =  — . 

2/ 


4.  r(„,-xi)sd.=£(5Li^'-5H?Lr:4^^LiM) 

J  oa^  16as 


iVote. — Let  X  ^^  a  sin'  9. 


APPENDIX  B. 


TABLE  OF  INTEGRALS. 

A  fairly  extensive  table  of  integrals  is  A  table  of  integrals,  arranged  by 
B.  O.  Peirce  and  published  by  Ginn  &  Co.,  Boston,  1890;  revised  edition  1899. 


Explanatory  Notes. 

To  say  that   j2x'dx   is   x^,   is  exactly  the  same  thing  as  to  say  that  the 

differential  of  x^  is  2x-dx. 

Constants  of  integration  are  omitted  from  this  table. 

Any  of  the  integrals  given  in  this  table  can  be  verified  by  differentia- 

— r-Tj  -\-  C)  according  to  Art.  30, 

— jjy  +  a  constant  J  is  the  integral  of 

x^'dx. 

sin  X 
One  has  frequent  occasion  to  use  the  following  formulas:  tan  x  = , 

cos  a;        1  1     J        1 

cot  X  =  - — ■,   sec  x  = ,  cosec  x  =  — — ,  and  vers  x  =   1  —  cos  x. 

sm  x'  cos  X  sm  x 

The  definitions  of  the  hyperbolic  functions  sinh  re,   cosh  x,   etc.,  are  given 

m  Chapter  VI,  Art.  94. 

The  letter  e  stands  for  the  base  of  the  Napierian  logarithms  ( =  2.7182818) ; 

log  X  stands  for  Napierian  logarithm  of  x. 


APPENDIX  B. 
A.  Fundamental  forms. — 

x"^  '  dx  = — j^   when  n  is  no%=-equal  to    —  1. 

3.    J  e*  •  dx  =  e^ 

a'  -  dx  —  i . 

log  a 

5.  j  sin  re  •  c^x  =  —  cos  x. 

6.  I  cos  X  '  dx  =  sin  a;. 

7.  J  tan  X  '  dx  =  —  log  cos  x. 

8.  I  cot  X  •  da;  =  log  sin  x. 

/dx 

0.  C  -  =  sinh~^  re  =  log  (re  +  Vl  -f  a^O* 
J   Vl  +  x^ 

1.  C  =  cosh-1  X  =  log  (x  +  Vx^  —  1). 

J    ^lx^  —  I 


dx 

4^x2 

dx         .     ^  ,  1.,     1+  X 


2.    J      T   o  =  tan~i  X, 


^- Jr^2  =  ^"^^"^  =  2^°^i-x 

J  xa/x^"^^ 


dx  , 

4.    I  — ,  =  sec~^  X 


24  APPENDIX  B. 

17.  I  — =  =  vers~i  a;. 

B.  Integrals  involving  trigonometric  functions.— 

18.  Jsccx  •  dx  =  log  (sec  a:  +  tan  x)  =  log  tan  f  ^  -f  |  j. 

19.  Jcosec  X  ^  dx  =  log  (cosec  x  ~  cot  x)  =  log  tan|. 

/x      1 
sin^  .T  •  da;  =  -  -  ^  sin  2a;. 

/x      1 
cos^  a;  •  c^a;  =  2  +  T  sin  2a;. 

22.  j    sec^  a;  •  rfa;  =  tan  a;. 

23.  J  cosec^  X  -  dx  =  —  cot  a;. 

24.  fsin  7.U;  sin  n^  .  dx  =  ^^^/^  "  ^^^  -  ^^^/^^^  +  ^,)^. 
J  2(w  -  n)  2(m  +  n) 

o<     To; .^  J  cos  (m  —  n)x      cos  (m  +  w)a; 

25.  I  sin  ma;  cos  na;  •  oa;  = ^ / ^ — -- — ~. 

J  2(m  —  n)  2(m  +  n) 

OA     /^^^^  J        sin  (m  —  n)a;  ,  sin  (m  +  n)a; 

26.  I  cos  wa;  cos  iia;  •  da;  =  —p-r ^  +  —7^7^ — ,     /  . 

•/  2(w  -  n)  2(w  +  n) 

27.  J  sin-i  a;  •  da;  =  a;  sin-^  x  +  Vl  -  x^. 

28.  J   cos~i  a;  •  da;  =  a;  cos"^  a;  —  Vl  —  a;^. 

C.  Integrals  involving  hyperbolic  functions. — 

29.  J  sinh  x  •  dx  =  cosh  a;. 


APPENDIX  B.  25 

30.  j  cosh  X  '  dx  =  sinh  x. 

31.  j  tanh  x  '  dx  =  log  cosh  x. 

32.  j  coth  X  '  dx  —  log  sinh  re. 

D.  Integrals  involving  (a  +  6a;). — 


J   (a 


+  6a; 
c?x  1 


(a  +  6x)2  6(a  +  hx) ' 

35.-  j~j^  =  p  [a  +  6a;  -  a  log  (a  +  6.t)]. 

E.  Integrals  involving  (a  -f  6a;2). — 

^^     C     dx  1  ,       ,  ic 

37.  J  — -r — „  =  -  tan-i  - 


38. 


c^  -\-  x'^      c  c 

dx  1  ,      c  -\-  X 


r     dx  1  , 


&  —  x^      2c       c  —  X 


-,«     r  ^  '  dx         1  ,      /  ,   ,  a\ 


F.  Integrals  involving     (a  +  6a;  +  ca;^). — For    brevity  let    X 
stand  for   (a  +  6a;  +  cx^)   and  let  g  stand  for   (4ac  —  6^). 

,^     rdx        2    .       ,2cx  +  h     ,  u.' 

40.    I  v^  =  — /^  tan~i ^ —  when  q  is  positive. 

4.1      C~—     ,        loe ; — -  when  o  is  negative. 

^^'  J  X        aT^     ^2ca;  +  6+A/-3 

G.  Integrals  involving  ^la  +  6a;. — 
42.  J  Va  +  6x  •  rfa;  =  25  '^(^  +  ^^)^* 


26 


APPENDIX  B. 


43.  fx^^+Tx  ■  &  =  -  ^(^^  -  ^^g/°  +  ^^)'. 
dx 


44 


1562 

_  2(8a2  -  12ahx  +  ISb^x^)  V(^TW' 
1056^ 


2Va  +  6i 


*  *^  Va  +  6a;  6 

r  a;  '  da;    ^  ^  2(2a  -  6x)  A/aT6x 
-^  Va  +  6x  362 

/^  x^'  dx    ^  2(8a^  -  4fl6a;  +  ^h^x^)  Va  +  6a; 

H.  Integrals  involving  Vo^^^^"^ — 

48.  C-ylaT^^^  '  dx  =  ~-<a^^^^  +  ^  sin-'-. 
J  2  2  a 

49.  fx-ylaF^"^^  •  dx  =  -  ^^i{a' -  x^y . 

50.  fx^  Va2  -  x2  •  (ia;  =  f  (2a;2  -  a^)  V^IT^a  4,  |'  gin-i  ? 
»/  o  o  a 

52.  r-p^= 

•^  A/a2--x2 


Va2  -  a;2       ,  ^-^ -_  ,      a  +  Va^  -  a;^ 

'  dx  =  -sa^  —  x^  —  a  log 


sm" 


Xj_^ 

Va2  -  a;2 
a;2  .  <?a; 


S3,    f-^1^-  =  _  V^ 


X" 


^.     r  x^  '  dx  X  /-; ^  ,    a^   .     ,  a; 

54.  I  —. =  -  Pi  Va2  -  a;2  +  -  sm-i  - . 

•^  Va2^^2  2  2  a 

55.  J'Co^  -  a;2)*  *  ^^  =  |  (5«'  "  ^a;^)  Va^  -  x"  + 


3a4  .     ,x 


<^ 


«/<.?^ 


x^)^       a^^a'-x^' 


APPENDIX  B.  27 

I'.  Integrals  involving  Va;^  +  a^. — 

57.  J'Vx2~+^2  '  dx  =  ^  A/iH-~a2  +  I'  log  {x  +  ViM^"S2). 

SS,  fx^¥^~a^  '  dx  =  ~^I'(^~+^K 

59.  J'a;^  Vx2~+~a2  •  da;  =  |  {2x^  +  a^)  Vi2Tp^2 

-  I  log  (x  +  V^+ a2). 

60..  f  ,   ^^        =  log  (a;  +  Va;^  +  a^). 
•-^  Va;^  +  a^ 

61.  r_Ai^=  V^+^2. 


62 


+ 
x^  '  dx        X    i-TT-; — i.     a^ 


.  f^^=  =  \  V:.^  +  a^-|-  log(x  +  V^H^a^). 
63.    r--^==i  log ^— =i  log  ^^^±^ 

'  -^  a;2  V^M^^  a2^ 

65.    r(a;2  +  a2)?  •  c?x  =  |(2a;2  +  5a^)  ^x^  +  a^ 


+  —  log  (a;  +  Va;2  +  a^). 


66     f        ^^        ^  ^ 

•  J   (a;2  +  a2)^      ^2  ^^.2  _|_  ^2" 


r'.  Integrals  involving  Vx^  -  a^.—AU  of  the  forms  57  to  66 
apply  in  this  case  by  simply  changing  the  sign  of  a^  except  form 
63  which  becomes: 

^^     r       dx  I         .X 

67.    I  — , =  -  sec-^  -. 

-^  x^^x^  -  a^       «  a 


28  APPENDIX  B. 


68. 


69.  rx>55irr^.<te=-3^±^z2^^g^^^r^^^^^^.5 

•/  D  J  a 

70.  f-,    ^     -=:vere-i^. 

71.  C-^LM==.  =  -  V2ax  -  x2  +  a  vers-i-*-. 
J    •V2aa;  -  a:^  a 

72.  r^^-^>  d:c  =  V2ax-x2  +  a  vers-^?. 
J  X  o, 

73.  f         ^  =  -    V2aa;  -  7^ 
J  X'^2ax  —  a?  ox 


APPENDIX  C 

A  SELECTED  LIST  OF  TREATISES  ON  MATHEMATICS  AND  ON 
THE  VARIOUS  BRANCHES  OF  MATHEMATICAL  PHYSICS. 

HISTORY. 

See  A  History  of  Mathematics  by  Florian  Cajori.  The  Macmillan 
Co.,  New  York,  1894 

A  good  sketch  of  the  history  of  calculus  is  given  in  the  article 
Infinitedmal  Calculus  in  the  9th  edition  of  the  Encyclopedia 
Britannica. 

Sir  David  Brewster's  Life  of  Newton j  Thomas  Constable  &  Co., 
Edinburgh,  1855,  is  well  worth  reading  by  anyone  who  is  inter- 
ested in  the  history  of  mathematics. 

The  theory  of  infinitesimals  and  the  theory  of  limits  have 
occasioned  a  very  great  deal  of  discussion  in  the  past.  These 
matters  are  discussed  somewhat  at  length  in  the  Britannica  article 
above  mentioned,  but  the  most  edifying  discussion  of  this  subject 
is  that  which  is  given  in  the  introductory  chapter  to  Augustus 
De  Morgan's  Differential  and  Integral  Calculus,  London,  1842. 

GENERAL  TREATISES. 

An  extremely  interesting  and  facinating  book  for  the  student  of 
mathematics  is  W.  K.  Clifford's*  Common  Sense  _oi  the  Ejcact 
Sciences,  The  International  Scientific  Series,  D.  Appleton  &  Co., 
New  York,  1888. 

One  of  the  best  general  treatises  is  Edouard_Gou.rsat's  Cot^rs 
d^ Analyse  Mathematique,  two  volumes,  Gauthier-Villars,  Paris, 
1905.     This  valuable  work  has  been  translated  into  English  by 

*W.  K.  Clifford's  philosophical  essays  (two  volumes,  Macmillan  &  Co., 
London,  1879)  and  especially  his  small  book  entitled  Seeing  and  Thinking 
(Macmillan  &  Co.,  London,  1880)  are  extremely  readable  and  interesting. 

29 


30  APPENDIX  C. 

E.  R.  Hedrick  and  it  is  published  as  Goursat-Hedrick  Mathe- 
matical  Analysis,  Ginn  &  Co.,  Boston. 

The  various  mathematical  articles  in  the  ninth  and  eleventh 
editions  of  the  Encyclopedia  Britannica  are  a  great  help  to  the 
student  of  mathematics. 

The  great  reference  work  in  pure  and  applied  mathematics  is 
the  Encyclopddie  der  mathematischen  Wissenschaften,  B.  G.  Teub- 
ner,  Leipsig.  The  publication  of  this  encyclopedia  was  begun  in 
1905  and  it  is  not  yet  finished.  In  the  meantime  the  work  is 
being  republished  in  a  revised  and  enlarged  form  in  a  French 
translation. 

The  most  extensive  general  treatise  on  mathematical  physics 
is  Winkelman^s  Handbuch  der  Physik  in  six  volumes,  second 
edition,  J.  A.  Earth,  Leipsig,  1909. 

THEORY  OF  FUNCTIONS. 

The  theory  of  functions  is  one  of  the  most  facinating  branches 
of  pure  mathematics. 

A  good  discussion  of  this  subject  is  given  in  the  second  volume 
of  Goursat's  Cours  d^ Analyse  Mathematique  (Goursat-Hedrick 
Mathematical  Analysis).  One  of  the  best  books  on  the  subject 
is  H.  Durdge's  Elements  of  the  Theory  of  Functions  (English  trans- 
lation by  Fisher  and  Schwatt),  Fisher  and  Schwatt,  Philadelphia, 
1896. 

Two  very  extensive  treatises  on  the  function  theory  are  A.  R. 
Forsythe's  Theory  of  Functions  of  a  Complex  Variable,  Cambridge 
University  Press,  1893;  and  Harkness  and  Morley^s  Theory  of 
Functions^  Macmillan  &  Co.,  London,  1893. 

DIFFERENTIAL  EQUATIONS 

A  very  good  elementary  treatise  is  D.  A.  Murray ^s  IntrodmAory 
Cours&JtL^DiJlerential  Equations,  Longmans  Green  &  Co.,  New 
York,  1897.  See  also  W.  W.  Johnson's  Differential  Equations^ 
John  Wiley  &  Sons,  New  York,  1890.    The  second  volume  of 


APPENDIX  C.  31 

Goursat-Hedrick's  Mathematical  Analysis  contains  a  good  dis- 
cussion of  differential  equations. 

A  very  good  treatise  on  partial  differential  equations  is  W.  E. 
Byerly^s  treatise  entitled  Fourier's  Series  and  Spherical,  Cylindrical 
and  Ellipsoidal  Harmonics,  Ginn  &  Co.,  Boston,  1893. 

An  important  general  method  of  solving  linear  differential 
equations  is  given  by  P.  A.  Lambert  in  the  Annals  of  Mathematics. 
First  paper  in  Vol.  XI,  pages  185-192,  July,  1910;  second  paper, 
Vol.  XIII,  pages  1-10,  September,  1911. 

QUATERNIONS  AND  VECTOR  ANALYSIS. 
(See  page  211.) 

HYPER-GEOMETRY  AND  RELATIVITY. 

A  very  interesting  development  of  mathematics  is  what  is 
called  non-Euclidean  geometry  or  hyper-geometry.  Perhaps  the 
most  interesting  thing  foi  a  beginner  to  read  on  this  subject  is 
Helmholtz's  popular  lecture  entitled  **Ueber  den  Ursprung  und 
die  Bedeutung  der  geometrischen  Axiome"  Vortrdge  und  Reden, 
Vol.  II,  F.  Vieweg  &  Son,  Braunschweig,  1884.  These  very  inter- 
esting lectures  of  Helmholtz  have  been  translated  into  English, 
first  series  by  Dr.  Pye-Smith,  second  series  by  E.  Atkinson,  and 
both  series  are  published  by  Longmans,  Green  &  Co. 

See  section  VI  of  article  Geometry  in  the  11th  edition  of  the 
Encyclopedia  Britannica.  A  good  discussion  of  non-Euclidean 
geometry  is  given  in  A.  N.  Whitehead's  Universal  Algebra,  Cam- 
bridge University  Press,  1898.  See  also  Stackel  und  Engel. 
TMm:ie_jderParallellinim  von  Euklid  his  auf  Gauss,  Leipsig.  1895; 
F.  Klein,  Nicht  Euklidische  Geometric,  Gottingen,  1893;  and  P. 
Barbarin,  La  Geometric  non-Euclidienne,  Paris,  1902. 

The  subject  of  non-Euclidean  geometry  is  closely  related  to  a 
very  recent  development  in  theoretical  physics  called  the  theory 
or  relativity.  A  very  simple  discussion  of  this  subject  is  given 
by  W.  S.  Franklin  in  an  article  entitled  The  Principle  of  Relativity, 
Journal  of  the  Franklin  Institute,  July,  1911.  A  very  interesting 
article  on  this  subject  is  the  eighth  lecture  in  Max  Planck's  Acht 


32  APPENDIX  C. 

Volemngen  uher  theoretische  Physik,  S.  Hirzel,  Leipsig,  1910.  The 
following  articles  by  Q,  N.  Lewis^  R.  C.  Tolman  and  E.  B.  Wilson^ 
are  interesting,  and  all  but  the  last  one  are  easy  to  read.  Philo- 
sophical Magazine,  Vol.  16,  pages  705-717;  American  Academy  Pro- 
ceedings, Vol.  44,  pages  711-724;  American  Academy  Proceedings, 
Vol.  48,  pages  389-507.  A  general  treatise  is  M.  Laue's  Das  Relativi- 
tdtsprimip,  Braunschweig,  F.  Vieweg  und  Sohn,  second  edition,  1913. 

ASTRONOMY.* 

Most  of  the  recent  researches  in  astronomy  have  been  in  the 
line  of  what  is  called  astro-physics,  and  but  few  additions  have 
been  made  to  the  older  theoretical  astronomy  with  its  beautifully 
complete  mathematical  theory. 

Good  books  on  theoretical  astronomy  for  the  beginner  are  C.  A. 
Young's  General  Astronomy,  Ginn  &  Co.,  Boston,  1898;  and  W.  W» 
Campbeirs  Elements  of  Practical  Astronomy,  The  Macmillan  Co., 
New  York,  1899. 

Perhaps  the  best  treatise  on  astronomical  measurements  is 
Wm..  Qhauvenet's  Syherical  and  Practical  Astronomy,  two  volumes, 
J.  B.  Lippincott,  Philadelphia,  1863.  For  a  treatise  on  astro- 
nomical calculations  see  Jas.  C.  Watson's  Theoretical  Astronomy , 
J.  B.  Lippincott,  Philadelphia,  1867.  See  also  ^ahnbestimmung 
der_  Kometen  und  Planeten,  T.  R.  v.  Oppolzer,  two  volumes,  Wm. 
Engelmann,  Leipsig,  1882. 

Perhaps  the  greatest  treatise  on  theoretical  astronomy  is  Traiti. 
d&.JLechanique  Cileste,  P.  S.  Laplace,  five  volumes,  Paris,  1799. 
English  translation  by  Nathaniel  Bowditch,  Boston,  1829. 

Theoretical  astronomy  goes  far  beyond  every  other  branch  of 

*  Although  we  are  here  concerned  chiefly  with  books  on  mathematical 
theory  it  is  worth  while,  perhaps,  to  mention  some  of  the  extremely  interesting 
descriptive  books  on  astronomy:  ^imon  Newcomb^s  Popular  Astronomy f 
Harper  Bros.,  New  York,  1882.  J.  Norman  Lockyer's  Stargazing,  Macmillan 
&  Co.,  London,  1878.  Richard  A.  Proctor's  Other  Worlds  than  Ours,  Longmans, 
Green  &  Co.,  London,  1878.  Also  C.  A.  Young's  General  Astronomy  is  ar- 
ranged so  that  the  general  reader  can  easily  omit  the  more  difficult  parts 
and  the  book  then  becomes  a  very  good  descriptive  treatise. 


APPENDIX  C.  33 

applied  mathematics  in  the  legitimate  use  of  elaborate  formulas 
and  in  the  tremendously  laborious  numerical  calculations  that 
are  involved.  The  student  can  get  an  idea  of  the  former  and  a 
faint  suggestion  of  the  latter  by  looking  over  the  four  large  volumes 
of  The  Collected  Mathematical  Works  of  George  W.  Hill  which  have 
been  recently  pubhshed  (1905)  by  The  Carnegie  Institution  of 
Washington  and  placed  in  every  important  library  in  the  United 
States.  Dr.  Hill's  greatest  contribution  to  theoretical  astronomy 
was  a  new  method  for  calculating  the  motion  of  the  moon. 

PROBABILITY. 

In  many  respects  the  theory  of  probability  is  the  most  important 
branch  of  mathematics  for  the  experimental  scientist. 

The  article  Prdbahility  in  the  eleventh  edition  of  the  Encyclopedia 
Britannica  is  a  very  good  outline  of  the  theory  of  probability. 
See  also  De  Morgan's  treatise  on  probabiHty  in  the  Cabinet 
Cyclopedia,  London,  1838.  Laplace's  Theorie  analytique  des 
Prohahilites^Vsins,  1820;  is  one  of  the  great  treatises  on  the  subject. 

The  application  of  the  theory  of  probability  to  measurement  is 
treated  in  Mansfield  Merriman's  Least  Squares,  John  Wiley  & 
Sons,  New  York,  eighth  edition,  1903;  in  R.  S.  Woodward's 
Prdbahility  and  Theory  of  Errors,  John  Wiley  &  Sons,  New  York, 
1906;  and  in  A.  de  F.  Palmer's  Theory  of  Measurements,  McGraw- 
Hill  Book  Co.,  New  York,  1912. 

The  application  of  the  theory  of  probability  to  the  study  of 
statistics  of  all  kinds  and  especially  to  biometrics  (the  quantitative 
study  of  variation  of  plants  and  animals)  is  of  great  importance. 
A  simple  discussion  of  this  subject  is  given  in  the  article  Variation 
and  Selection  in  the  eleventh  edition  of  the  Encyclopedia  Britannica. 

The  kinetic  theory  of  gases,  one  of  the  most  important  branches 
of  theoretical  physics  is  a  phase  of  the  theory  of  probability.  See 
Watson's  Kinetic  Theory  of  Gases,  The  Clarendon  Press,  Oxford, 
1893.  See  also  the  epoch-making  work,  T^dwig  "Rn1t7mfl.Tm\q 
Ymksungm^uh^.GaMMQrie^iT^  two  parts,  J.  A.  Barth,  Leipsig, 
1895  and  1898.    The  kinetic  theory  of  gases  is  treated  in  a  very 


34  APPENDIX  C. 

general  way  by  J.  Willard  Gibbs  in  his  Statistical  MechanicSf 
Charles  Scribner's  Sons,  New  York,  1902. 

A  very  good  discussion  of  the  kinetic  theory  of  gases  is  given  in 
Nernst's  Theoretical  Chemistry,  Chapter  II,  translated  by  Lehfeldt, 
Macmillan  &  Co.,  London,  1904. 

The  ideas  of  the  kinetic  theory  of  gases  are  extensively  used  in 
the  recent  theories  of  the  discharge  of  electricity  through  gases 
and  in  the  theories  of  radioactivity.  See  especially  J.  J.  Thomson's 
Conduction  of  Electricity  through  Gases,  Cambridge  University 
Press,  1906.  See  also  E.  Rutherford's  Radioactive  Transforma- 
tions, Charles  Scribner's  Sons,  New  York,  1906. 

The  modem  statistical  theory  of  radiant  heat  is  also  a  branch 
of  the  theory  of  probability.  Some  idea  of  this  subject  can  be 
obtained  from  lecture  6  of  Acht  Vorlesungen  iiher  theoretische 
Physik,  Max  Planck,  S.  Hirgel,  Leipsig,  1910. 

THERMODYNAMICS. 

There  is  a  widespread  notion  that  theoretical  thermodynamics 
makes  a  very  severe  demand  upon  the  methods  of  higher  mathe- 
matics, whereas,  as  a  matter  of  fact  its  demands  are  less  perhaps 
than  any  other  branch  of  mathematical  physics. 

An  extremely  simple  development  of  the  mathematical  theory 
of  thermodynamics  is  given  in  Franklin  and  MacNutt's  Mechanics 
and  Heat,  pages  350-397,  The  Macmillan  Co.,  1910. 

A  very  good  advanced  treatise  is  Max  Planck's  Treatise  on 
Thermodynamics  (English  translation  by  Alexander  Ogg),  Long- 
mans Green  &  Co.,  London,  1903.  Another  important  treatise  is 
Edgar  Buckingham's  Theory  of  Thermodynamics ^  The  Macmillan 
Co.,  New  York,  1900. 

R.  Clausius'  Mechanische  Wdrmetheorie,  third  edition,  F. 
Vieweg  &  Sons,  Braunschweig,  1887,  is  a  very  important  work. 

The  student  interested  in  theoretical  thermodynamics  will  find 
it  well  worth  while  to  read  the  celebrated  paper  of  J.  Van't  Hoff 
entitled  "The  function  of  osmotic  pressure  in  the  analogy  between 
solutions  and  gases,"  Philosophical  Magazine,  Vol.  XXVI,  pages 
81-105,  August,  1888. 


APPENDIX  C.  35 

The  best  treatment  of  thermodynamics  for  the  experimental 
chemist  is  that  which  is  given  in  a  more  or  less  disconnected 
manner  in  W.  Nernst's  Theoretical  Chemistry,  English  translation 
by  R.  A.  Lehfeldt,  Macmillan  &  Co.,  London,  1904. 

A  good  treatment  of  thermodynamics  for  the  steam  engineer 
is  that  which  is  given  in  chapters  II  and  III  (pages  37-159)  of 
JL_A,_-Ewing'  s  The  Steam  Engine  _  and  other  Heat  Engines^  third 
edition,  Cambridge  University  Press,  1910. 

THEORETICAL  MECHANICS. 

.  There  are  many  good  beginner's  treatises  on  theoretical  me- 
chanics. A  good  advanced  treatise  is  Alexander  Ziwet's  Ele- 
mentary Treatise  on  Theoretical  Mechanics,  in  two  parts,  The  Mac- 
millan Co.,  New  York,  1893. 

Every  student  of  mathematical  physics  should  read  a  portion, 
at  least,  of  Sir  Isaac  Newton's  Prindpia. 

An  extremely  interesting  and  readable  book  is  Poinsot's  Theorie 
Nouvelle  de  la  Rotation  des  Corps,  second  edition,  Paris,  1851. 
Harold  Crabtree's  Spinning  Tops,  Longmans  Green  &  Co., 
London,  1909  (about),  is  a  good  mathematical  treatment  of  the 
theory  of  the  rotation  of  a  rigid  body. 

The  most  exhaustive  treatises  on  theoretical  mechanics  are; 
Geo.  M.  Minchin's  Treatise  on  Statics,  third  edition,  two  volumes. 
The  Clarendon  Press,  Oxford,  1886;  Edward  J.  Routh's  Ele- 
mentary Rigid  Dynamics,  Macmillan  &  Co.,  London,  1860;  and 
Edward  J.  Routh's  Advanced  Rigid  Dynamics,  Macmillan  &  Co., 
London,  1860. 

See  lecture  7  in  Acht  Vorlesungen  uber  theoretische  Physik  by 
Max  Planck,  S.  Hirzel,  Leipsig,  1910.  This  lecture  deals  with  the 
most  fundamental  principle  of  physics,  the  so-called  principle  of 
least  action. 

THEORY  OF  SOUND.  ♦ 

One  of  the  best  books  on  this  subject  for  the  beginner  is  J.  H. 
Poynting  and  J.  J.  Thomson's  Sound,  Charles  Griffin  &  Co., 
London,  1899.    See  also  the  article  Acoustics  in  the  9th  edition 


36  APPENDIX  C. 

of  the  Encyclopedia  Britannica.  Helmholtz^s  great  work  The 
Sensations  of  Tone  (English  translation  by  Alexander  J.  Ellis), 
contains  a  series  of  appendices  on  mathematical  theory. 

The  most  comprehensive  treatise  is  Lord  Rayleigh^s  Theory  of 
Sound  in  two  volumes,  Macmillan  &  Co.,  London,  1877.  Second 
edition  revised  and  enlarged  1894. 

THEORY  OF  ELECTRICITY  AND  MAGNETISM. 

An  understanding  of  vector  analysis,  as  developed  in  chapter  IX 
of  this  text,  is  absolutely  necessary  before  one  can  begin  the 
study  of  Maxwell's  theory  of  electricity  and  magnetism. 

A  very  simple  discussion  of  Maxwell's  theory  is  given  in  Frank- 
lin's Electric  Waves,  pages  186-196,  The  Macmillan  Co.,  New 
York,  1909.  The  beginner  will  be  greatly  helped  by  E.  Atkinson's 
translation  of  Mascart  and  Joubert's  Treatise  on  Electricity  and 
Magnetism,  Vol.  I,  Thos.  de  la  Rue  &  Co.,  London,  1883. 

The  great  treatise  on  this  subject  is  Maxwell's  original  treatise 
in  two  volumes  entitled  Electricity  and  Magnetism.  The  first 
edition  of  this  work  appeared  in  1873.  Third,  edition  very 
slightly  altered,  The  Clarendon  Press,  1891. 

The  study  of  Maxwell's  treatise  is  greatly  facilitated  by  Mascart 
and  Joubert's  treatise  above  mentioned ;  by  the  study  of  Heinrich 
Hertz's  Electric  Waves,  translated  by  D.  E.  Jones,  Macmillan  & 
Co.,  London,  1893;  and  by  the  study  of  A.  G.  Webster's  Electricity 
and  Magnetism,  Macmillan  &  Co.,  London,  1897. 

A  most  excellent  treatise  for  the  student  is  Abraham  and 
Foppl's  Theorie  der  Ehctrizitdt;  Vol.  I,  Einfuhrung  in  die  Max- 
wellsche  Theorie,  3d  edition,  Leipsig,  1907;  Vol.  II,  Electromag- 
neschie  Theorie  der  Strahlung  (Electronentheorie) ,  Leipsig,  1905. 

Recent  Researches  Electricity  and  Magnetism  by  J.  J.  Thomson, 
The  Clarendon  Press,  Oxford,  1893,  contains  a  great  deal  of 
interest  especially  on  the  subject  of  electric  waves.  See  also  J.  J. 
Thomson's  _  Cowc^t^c^ion  qf_  Electricity  through  Gases,  Cambridge 
University  Press,  1906.  This  important  book  deals  principally 
with  the  electron  theory,  although,  of  course,  the  book  deals 
almost  entirely  with  experimental  researches. 


APPENDIX  C.  37 

THE  THEORY  OF  LIGHT. 

One  of  the  best  books  for  the  student  is  Thomas  Preston's 
Theory  of  Lights  third  edition,  Macmillan  &  Co.,  London,  1901. 

An  extremely  interesting  book,  partly  theoretical  and  partly 
descriptive,  is  A.  A.  Michelson's  Light  Waves  and  their  UseSf  Uni- 
versity of  Chicago  Press,  1903.  R.  W.  Wood's  PhydcaLQpiic&f 
second  edition.  The  Macmillan  Co.,  New  York,  1910,  contains  a 
great  deal  of  interesting  and  important  theory. 

The  theory  of  lenses  and  optical  instruments  is  treated  in  a 
very  exhaustive  manner  by  Czapski,  von  Rohr  and  Eppfenstein 
in  Winkelmann's  Handbuch  der  Phydk. 

One  of  the  most  interesting  series  of  original  memoirs  is  that 
of  Augustin  Fresnel,  see  Fresnel's  Oevres  Completes^  Vol.  I,  pages 
1-382,  Paris,  1866. 

A  very  complete  treatise  on  the  theory  of  light  (electro- 
magnetic) is  that  of  P.  Prude;  Enghsh  translation  by  C.  R.  Mann 
and  R.  A.  MiUikan,  Longmans  Green  &  Co.,  New  York,  1902. 

HYDROMECHANICS. 

One  of  the  best  treatises  on  this  subject  for  the  student  is  the 
article  Hydromechanics  in  the  ninth  edition  of  the  Encyclopedia 
Britannica.  Part  I  of  this  article  is  devoted  to  Hydrostatics.  In 
this  part  the  problem  of  the  figure  of  the  earth  is  discussed  and 
also  the  important  practical  problem  of  the  stability  of  floating 
bodies.  Part  II  of  this  article  is  devoted  to  Hydrodynamics,  the 
highly  mathematical  theory  of  the  motion  of  a  frictionless  fluid. 
Part  III  of  this  article  is  devoted  to  Hydraulics  from  the  point  of 
view  of  the  experimental  physicist  and  the  engineer.  Parts 
I  and  II  were  written  by  A.  G.  Greenhill  and  Part  III  was  written 
by  W.  C.  Unwin.  Professor  Unwin's  article  has  been  pubUshed 
as  a  separate  treatise,  Macmillan  &  Co.,  London,  1907. 

See  also  G.  M.  Minchin's  Treatise  on  Hydrostatics,  two  volumes, 

second  edition,  Clarendon  Press,  Oxford,  1912;   Horace  Lamb's 

Hydrodynamics  I  Cambridge  University  Press,  1879;  third  edition 

1906;  and  A.  B.  Bassett's  Hydrodynamics,  two  volumes,  Deighton 

Bell  &  Co.,  Cambridge,  1888. 
20 


38  APPENDIX  C. 

THEORY  OF  ELASTICITY. 

An  understanding  of  vector  analysis  as  developed  in  chapter  IX 
of  this  text  is  very  helpful  in  the  study  of  the  theory  of  elasticity. 
Thus  the  ideas  which  are  established  in  Art.  136  are  the  foundation 
of  the  elementary  theory  of  elasticity  as  given  ii;  Franklin  and 
MacNutt's  Mechanics  and  Heat,  pages  182-218,  The  Macmillan 
Co.,  New  York,  1910.  This  is  perhaps  the  simplest  existing 
elementary  treatise  on  the  theory  of  elasticity. 

The  matter  which  is  discussed  in  Art.  136,  namely,  the  theory 
of  elastic  strains,  is  discussed  in  W.  K.  Clifford's  Kinematic, 
part  I,  Macmillan  &  Co.,  London,  1878,  pages  158-221. 

One  of  the  best  treatises  on  the  theory  of  elasticity  is  W.  J. 
Ibbetson's  Mathematical  Theory  of  Elasticity,  Macmillan  &  Co., 
London,  1887;  see  also  A.  E.  H.  Love's  Mathematical  Theory  of 
Elasticity,  two  volumes,  Cambridge  University  Press,  1892. 

The  greatest  reference  work  in  the  theory  of  elasticity  is  Karl 
Pearson's  History. 

CRYSTALLOGRAPHY. 

It  is  not  generally  known  that  one  of  the  most  interesting  and 
remarkable  branches  of  mathematical  physics  is  the  theory  of 
crystallography.  Thus  the  purely  mathematical  theory  of  regu- 
lar-point-systems in  space  is  in  complete  accord  with  experi- 
mental studies  of  crystal  forms.  A  regular-point-system  in  space 
is  called  a  space  lattice.  The  purely  mathematical  theory  of  the 
space  lattice  is  treated  in  Sohncke's  Theorie  der  Krystallstruktur, 
B.  G.  Tuebner,  Leipsig,  1879.  Very  complete  treatises  on 
crystallography  are  Groth's  Physikalische  Krystallographie,  Wm. 
Englemann,  Leipsig,  1895;  and  Liebisch's  Grundriss  der  Physi- 
kalischen  Krystallographie,  Veit  &  Co.,  Leipsig,  1896.  In  both  of 
these  books  the  geometrical  theory  of  crystallography  is  fully 
treated. 


INDEX 


Acceleration  and  velocity,  55 

in  circular  motion,  70 
Algebraic  integration,  82 
Angular  acceleration  and  torque,  146 
Arbitrary  variations,  9 
Area  under  a  curve,  28 
Artificial  functions  and  natural  func- 
tions, 15 
Average,  discussion  of,  121 

value  of  a  function,  121 

Barrel  hoop,  discussion  of,  72 
Beam,  the  problem  of  the  bent,  116 
Bent  beam,  problem  of,  115 
Boat,  starting  of  a,  177 
stopping  of  a,  178 

Center  of  gravity,  123 

of  mass.     See  center  of  gravity, 
of  pressure,  134 
Change,  rate  of,  2 
Circle,  the  osculating,  66 
Complex  quantity,  definition  of,  163 
geometrical     representation 

of,  163 
use  of,  153 
Component     slopes     and     resultant 

slopes,  96 
Constant  of  integration,  33 
Constant    quantities    and    variable 

quantities,  1 
Continuous    variables    and    discon- 
tinuous variables,  4 
Convergence  of  a  vector  field,  229 
Convergent  series,  43 
Cosines,  hyperbolic,  166 
Curl,  cartesian  expression  for,  238 
divergence  of,  240 
example  of,  237 
of  a  gradient,  239 
of  a  vector  field,  236 
Curvature,  61 

radius  of,  65 
Cyclometer,  the,  74 

Dam,  force  exerted  upon,  136 


Damped  oscillations,  185 
Decrements  and  increments,  2 
Demoivre's  theorem,  162 
Dependent  variables,  13 
Derivative,  Graphical  representation 
of  a,  18 
of  a  function,  17 
Derivatives,  successive,  54 
Definite  integrals  and  indefinite  in- 
tegrals, 34 
Differential  equation,  31  and  168 
Degree  and  order  of,  168 
general   solutions   and   par- 
ticular solutions  of,  172 
linear,  172 
of  wave  motion,  190 
ordinary  and  partial,  168 
partial,  89 
pure  and  mixed,  169 
Differentiation  and  integration,  32 
formulas  for,  38 
by  development,  38 
by  rule,  38 
geometric,  70 
graphical,  38 

of  an  exponential  function,  48 
of  a  function  of  a  function,  42 
of  an  implicit  function,  91 
of  a  logarithm,  46 
of  a  product,  40 
of  a  quotient,  41 
of  a  sum,  39 

of  trigonometric  functions,  50 
partial,  87 

successive,  88 
physical,  38 
rules  for,  38-54 
successive,  54 
Discontinuous    variables    and    con- 
tinuous variables,  4 
Distributed  scalar,  218 

vector,  218 
Divergence,  Cartesian  expression  for, 
229 
of  a  curl,  240 
of  a  vector  field,  228 


40 


INDEX. 


Equation,  differential,  partial,  89 
Errors  of  observation,  influence  upon 

a  result,  109 
Euler's  expression  for  sine  and  cosine, 

164 
Expansions  in  series,  153 

Fields,  scalar  and  vector,  217,  218 
Forced  oscillations,  188 
Fourier's  theorem,  199 

application  of,  204 
Function,  definition  of,  13 

derivative  of  a,  17 

graphical  representation  of  a,  16 

implicit  differentiation  of,  91 

law  of  growth  of,  31 

tabulation  of,  15 
Functions,  natural  and  artificial,  15 

which  have  the  same  derivative, 
25 

Gravity,  center  of,  123 
Geometric  differentiation,  70 
Grade,  definition  of,  6 
Gradient,  curl  of,  239 

definition  of,  6 

line  integral  of,  223 

of  a  distributed  scalar,  219 

of  a  hill,  93 

temperature,  discussion  of,  7 
Gradients,  examples  of,  98 
Graphical  representation  of  a  deriva- 
tive, 18 
of  a  function,  16 
Gyration,  radius  of,  142 

Harmonic  motion,  58 
Hill,  gradient  of  a,  93 

slope  of  a,  93 

surface,  area  of,  105 

volume  of  a,  lOO 
Hooke's  law,  115 
Hoop,  discussion  of,  72 
HyperboUc  sines  and  cosines,  166 

Implicit  function,  differentiation  of, 

91 
Increments  and  decrements,  2 
Indefinite  integrals  and  definite  inte- 
grals, 34 
Independent  variables,  13 
Inertia,  moment  of.  140 

moments  of,  about  parallel  axes, 
147 


Inertia,  moments  of,  table  of,  151 
Infinitesimals,  method  of,  22 
Integrals,  definite  and  indefinite,  34 

table  of.     See  Appendix  B. 
Integrating  machine,  74 
Integration,  74 

algebraic,  82 

and  differentiation,  32 
formulas  for,  38 

by  rules.    See  footnote,  page  83. 

by  steps,  79 

constant  of,  33 

mechanical,  74 

multiple.    See  Integration,  Par- 
tial. 

partial,  87  and  99 

rules  for,  84 
Irrotational  vector  fields,  242 

Kelvm's  law,  113 

Line  integral,   Cartesian   expression 
for,  223 
of  a  gradient,  223 
of  a  vector  field,  222 
reduction  of  to  a  surface  in- 
tegral, 239 
Linear  vector  field,  the,  243 

Maclaurin's  theorem,  153 

appHed  to  the  function  of 
two  variables,  159 
Mass,    center    of.     See    Center    of 

gravity. 
Maximum  and  minimum  values,  111 
Mechanical  integration,  74 
Median  line  of  beam,  definition  of,  117 
Method  of  infinitesimal,  the,  22 
Minimum  and  maximum  values.  111 
Moment  of  inertia,  140 

of  section  of  a  beam,  148 
of  inertia  about  parallel  axes,  147 
Table  of,  151 
Motion,  harmonic,  58 

in  a  circle,  70 
Multiple    integration.     See    Partial 
Integration. 

Natural  functions  and  artificial  func- 
tions, 15 

Oscillations,  damped,  185 

forced,  188 

undamped,  181 
Osculating  circle,  66 


INDEX. 


41 


Partial  differential  equation,  89 

differentiation,  87 
successive,  88 

integration,  87  and  99 
Planimeter,  the,  74 
Plucked  string,  the  vibration  of,  197 
Potential,  existence  theorem  of,  225 

multivalued,  226 

of  a  vector  field,  225 

scalar,  225 

and  vector,  242 
Pressure,  center  of,  134 
Principle  of  superposition,  174 
Prismoid  formula,  the,  102 

Quantities,  constant  and  variable,  1 

Radius  of  curvature,  65 

of  gyration,  142 
Rate  of  change,  2 
Resultant    slopes    and     component 

slopes,  96 
Rotational  vector  fields,  242 

Scalar  and  vector  quantities,  211 
field,  217 

gradient  of,  219 
volume  integral  of,  219 
function,  218 
potential,  225 

and  vector  potential,  242 
Series,  convergent,  43 
Sines,  hyperbolic,  166 
Slope  of  a  hill,  93 

Slopes,  component  and  resultant,  96 
Solenoidal  vector  fields,  235 
Space  analysis,  210 
Speed-time  curve,  the,  79 
Spring,  work  required  to  stretch  a,  26 
Starting  of  a  boat,  177 
Stopping  of  a  boat,  178 
Stream  lines,  221 
Stretched  string,  equation  of  motion 

of,  193 
String,  equation  of  motion  of,  193 

plucked,  vibration  of,  197 
Successive  derivatives,  54 
differentiation,  54 
partial  differentiation,  88 
Superposition,  the  principle  of,  174 
Surface  integral,  Cartesian  expression 
for,  228 
of  distributed  vector,  227 


Surface  integral,  reduction  of  to  line 
mtegral,  239 
of  to  volume  integral, 
234 
of  a  hill,  area  of,  105 

Table  of  moments  of  inertia,  151 
Tabulation  of  a  function,  15 
Tangent  plane,  equation  of,  94 
Taylor's  theorem,  157 
Temperature  gradient,  discussion  of,  7 
Torque  and  angular  acceleration,  146 
Tube  of  flow  in  a  vector  field,  235 

Undamped  oscillations,  181 
Unit  tube  in  a  vector  field,  235 

Variable  and  constant  quantities,  1 
Variables,    continuous    and    discon- 
tinuous, 4 
dependent  and  independent,  13 
Variations,  arbitrary,  9 
Vibrations,  damped,  185 
forced,  188 
undamped,  181 
Vector  analysis,  210 

and  scalar  quantities,  211 
field,  218 

convergence  of,  229 
curl  of,  236 
divergence  of,  228 
line  integral  of,  222 
potential  of,  225 
rotational  and  irrotational, 

242 
solenoidal,  235 
surface  integral  of,  227 
the  linear,  243 
function,  218 
potential,  241 

and  scalar  potential,  242 
products,  213 
Velocity  and  acceleration,  55 
Volume  integral  of  scalar  field,  219 

reduction  of  to  surface  in- 
tegral, 234 
of  a  hill,  100 
Vortex  motion,  241 

Water  gate,  force  exerted  upon,  134 
Watt-hour-meter,  the,  74 
Wave  motion,  differential  equation  of, 
190 


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